Toisen kertaluvun differentiaaliyhtälöt ovat differentiaaliyhtälöitä, joissa esiintyy funktioiden toisen kertaluvun derivaattoja. Toisen kertaluvun DY:t ovat yleisesti ottaen vaikeampia ratkaista kuin ensimmäisen kertaluvun. Toisen kertaluvun DY on funktio, joka on muotoa
F
(
t
,
x
,
x
′
,
x
″
)
=
0
{\displaystyle F(t,x,x',x'')=0}
DY on normaalimuotoinen , jos se on muotoa
x
″
=
f
(
t
,
x
,
x
′
)
{\displaystyle x''=f(t,x,x')}
Yksinkertaisuuden vuoksi rajoitetaan tämän aihealueen tarkastelut vain normaalimuotoisiin ja lineaarisiin DY:ihin . Toisen kertaluvun DY:iden ratkaiseminen eroaa ensimmäisen kertaluvun DY:iden ratkomisesta mm. seuraavasti:
Yleistä ratkaisukaavaa ei ole.
Jokaiselle yhtälötyypille on oma teoriansa.
Ratkaisuissa yritetään löytää ns. perusratkaisut , joiden avulla kaikki ratkaisut voidaan muodostaa. Ratkaise differentiaaliyhtälö
F
(
t
,
x
,
x
′
,
x
″
)
=
0
{\displaystyle F(t,x,x',x'')=0}
kaikki derivoituvat funktiot
x
(
t
)
{\displaystyle x(t)}
yleistä yhtälöä, ei erikoistapausta .Ratkaise alkuarvotehtävä
{
F
(
t
,
x
,
x
′
,
x
″
)
=
0
x
(
t
0
)
=
x
0
x
′
(
t
0
)
=
v
0
{\displaystyle {\begin{cases}F(t,x,x',x'')=0\\x(t_{0})=x_{0}\\x'(t_{0})=v_{0}\end{cases}}}
tarkoittaa sitä, että on etsittävä jokin derivoituva funktio
x
:
]
x
0
−
δ
,
x
0
+
δ
[
→
R
{\displaystyle x:\,]x_{0}-\delta ,x_{0}+\delta [\to \mathbb {R} }
δ
>
0
{\displaystyle \delta >0}
x
(
t
0
)
=
x
0
{\displaystyle x(t_{0})=x_{0}}
x
(
t
1
)
=
x
1
{\displaystyle x(t_{1})=x_{1}}
reuna-arvotehtäväksi . Määritelmä 1: lineaarinen differentiaaliyhtälö
muokkaa
Toisen kertaluvun normaalimuotoinen DY on lineaarinen , jos se on muotoa
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
(
t
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q(t)}
missä
p
,
r
,
q
:
]
a
,
b
[
→
R
{\displaystyle p,r,q:\,]a,b[\to \mathbb {R} }
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
a
<
b
{\displaystyle a<b}
x
{\displaystyle x}
t
{\displaystyle t}
Jos toisen kertaluvun normaalimuotoinen differentiaaliyhtälö on muotoa
x
″
=
f
(
t
,
x
′
)
{\displaystyle x''=f(t,x')}
päästään ratkaisun kimpuun käyttämällä apufunktiota
y
(
t
)
=
x
′
(
t
)
{\displaystyle y(t)=x'(t)}
y
{\displaystyle y}
y
′
=
f
(
t
,
y
)
{\displaystyle y'=f(t,y)}
joka on ensimmäisen kertaluvun DY ja siten ratkaistavissa aiemmin opituin menetelmin. Kun
y
{\displaystyle y}
x
=
∫
y
d
t
+
C
{\displaystyle x=\int y\,\mathrm {d} t+C}
missä
C
{\displaystyle C}
Määritelmä 2: homogeeninen differentiaaliyhtälö
muokkaa
Toisen kertaluvun lineaarinen DY on homogeeninen , jos
q
(
t
)
≡
0
{\displaystyle q(t)\equiv 0}
Kuuluisia differentiaaliyhtälöitä
muokkaa
Toisen kertaluvun differentiaaliyhtälöt ovat erityisesti siitä tärkeitä, että niillä on usein jokin sovellus tai merkitys fysiikassa . Niiden ratkaisut eivät myöskään yleensä ole eksplisiittisiä. Joidenkin DY:iden ratkaisufunktiot ja itse DY:t on nimetty keksijänsä mukaan. Tässä on niistä joitain.
Legendren differentiaaliyhtälö
muokkaa
Differentiaaliyhtälöitä, jotka ovat muotoa
(
1
−
t
2
)
x
″
−
2
t
x
′
+
λ
(
λ
+
1
)
x
=
0
{\displaystyle (1-t^{2})x''-2tx'+\lambda (\lambda +1)x=0}
missä
λ
∈
R
{\displaystyle \lambda \in \mathbb {R} }
Legendren differentiaaliyhtälöiksi
λ
∈
N
{\displaystyle \lambda \in \mathbb {N} }
Legendren polynomeja .
Besselin differentiaaliyhtälö
muokkaa
Differentiaaliyhtälöitä, jotka ovat muotoa
t
2
x
″
+
t
x
′
+
(
t
2
−
μ
2
)
x
=
0
{\displaystyle t^{2}x''+tx'+(t^{2}-\mu ^{2})x=0}
missä
μ
∈
R
{\displaystyle \mu \in \mathbb {R} }
Besselin differentiaaliyhtälöiksi
μ
∈
N
{\displaystyle \mu \in \mathbb {N} }
Besselin funktioita .
Hypergeometrinen yhtälö
muokkaa
Differentiaaliyhtälöitä, jotka ovat muotoa
t
(
1
−
t
)
x
″
+
(
c
−
(
a
+
b
+
1
)
t
)
x
′
−
a
b
x
=
0
{\displaystyle t(1-t)x''+\left(c-(a+b+1)t\right)x'-abx=0}
missä
a
,
b
,
c
∈
R
{\displaystyle a,b,c\in \mathbb {R} }
hypergeometrisiksi differentiaaliyhtälöiksi . Näiden DY:iden ratkaisut ovat nk. hypergeometrisia funktioita .
Homogeenisen differentiaaliyhtälön ratkaiseminen
muokkaa
Olkoon funktiot
x
1
(
t
)
{\displaystyle x_{1}(t)}
x
2
(
t
)
{\displaystyle x_{2}(t)}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
ratkaisuja. Tällöin myös niiden lineaarikombinaatio
w
(
t
)
=
α
x
1
(
t
)
+
β
x
2
(
t
)
{\displaystyle w(t)=\alpha x_{1}(t)+\beta x_{2}(t)}
missä
α
,
β
∈
R
{\displaystyle \alpha ,\beta \in \mathbb {R} }
Määritelmä 3: ratkaisukanta
muokkaa
Funktiot
x
1
,
x
2
∈
C
2
(
]
a
,
b
[
)
{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{2}(]a,b[)}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
ratkaisukannan , jos jokaiselle saman DY:n ratkaisulle
x
{\displaystyle x}
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
{\displaystyle x(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Ratkaisukanta on siis eräänlainen DY:n perusratkaisuiden joukko, jonka avulla voidaan esittää saman DY:n mikä tahansa muu ratkaisu. Jotta ratkaisukanta
{
x
1
,
x
2
}
{\displaystyle \{x_{1},x_{2}\}}
Funktioiden
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Funktioiden
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
″
=
−
x
{\displaystyle x''=-x}
x
1
(
t
)
=
sin
t
{\displaystyle x_{1}(t)=\sin t}
x
2
(
t
)
=
2
sin
t
{\displaystyle x_{2}(t)=2\sin t}
x
3
(
t
)
=
cos
t
{\displaystyle x_{3}(t)=\cos t}
{
x
1
,
x
2
}
{\displaystyle \{x_{1},x_{2}\}}
x
3
{\displaystyle x_{3}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
{
x
1
,
x
3
}
{\displaystyle \{x_{1},x_{3}\}}
Funktioiden
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Määritelmä 4: lineaarinen riippuvuus ja riippumattomuus
muokkaa
Funktiot
f
,
g
:
]
a
,
b
[
→
R
{\displaystyle f,g:\,]a,b[\,\to \mathbb {R} }
lineaarisesti riippuvia (lyhennetään LD , en. L inearly D ependent
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
c
1
≠
0
{\displaystyle c_{1}\neq 0}
c
2
≠
0
{\displaystyle c_{2}\neq 0}
c
1
f
(
t
)
+
c
2
g
(
t
)
=
0
{\displaystyle c_{1}f(t)+c_{2}g(t)=0}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
c
∈
R
{\displaystyle c\in \mathbb {R} }
f
(
t
)
=
c
g
(
t
)
{\displaystyle f(t)=cg(t)}
g
(
t
)
=
c
f
(
t
)
{\displaystyle g(t)=cf(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
.
Vastaavasti funktiot
f
,
g
:
]
a
,
b
[
→
R
{\displaystyle f,g:\,]a,b[\,\to \mathbb {R} }
lineaarisesti riippumattomia (lyhennetään LI , en. L inearly I ndependent
c
1
f
(
t
)
+
c
2
g
(
t
)
=
0
{\displaystyle c_{1}f(t)+c_{2}g(t)=0}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
seuraa, että
c
1
=
c
2
=
0
{\displaystyle c_{1}=c_{2}=0}
HUOM! Jos
f
,
g
:
]
a
,
b
[
→
R
{\displaystyle f,g:\,]a,b[\,\to \mathbb {R} }
g
(
t
)
≡
0
{\displaystyle g(t)\equiv 0}
f
{\displaystyle f}
g
{\displaystyle g}
[1]
Osoitetaan, että funktiot
f
,
g
:
R
→
R
{\displaystyle f,g:\mathbb {R} \to \mathbb {R} }
f
(
t
)
=
e
t
{\displaystyle f(t)=\mathrm {e} ^{t}}
g
(
t
)
=
e
2
t
{\displaystyle g(t)=\mathrm {e} ^{2t}}
Antiteesi [2]
f
{\displaystyle f}
g
{\displaystyle g}
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
c
1
≠
0
{\displaystyle c_{1}\neq 0}
c
2
≠
0
{\displaystyle c_{2}\neq 0}
c
1
f
(
t
)
+
c
2
g
(
t
)
=
0
{\displaystyle c_{1}f(t)+c_{2}g(t)=0}
kaikilla
t
∈
R
{\displaystyle t\in \mathbb {R} }
c
1
e
t
+
c
2
e
2
t
=
0
e
t
(
c
1
+
c
2
e
t
)
=
0
|
|
:
e
t
(
≠
0
∀
t
∈
R
)
c
1
+
c
2
e
t
=
0
c
2
e
t
=
−
c
1
{\displaystyle {\begin{aligned}c_{1}\mathrm {e} ^{t}+c_{2}\mathrm {e} ^{2t}&=0\\\mathrm {e} ^{t}(c_{1}+c_{2}\mathrm {e} ^{t})&=0\qquad ||:\mathrm {e} ^{t}(\neq 0\,\forall \,t\in \mathbb {R} )\\c_{1}+c_{2}\mathrm {e} ^{t}&=0\\c_{2}\mathrm {e} ^{t}&=-c_{1}\end{aligned}}}
Derivoidaan yhtälö puolittain:
d
d
t
c
2
e
t
=
d
d
t
(
−
c
1
)
c
2
e
t
=
0
{\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}c_{2}\mathrm {e} ^{t}&={\frac {\mathrm {d} }{\mathrm {d} t}}(-c_{1})\\c_{2}\mathrm {e} ^{t}&=0\end{aligned}}}
Tämän pitää päteä kaikilla
t
∈
R
{\displaystyle t\in \mathbb {R} }
c
2
=
0
{\displaystyle c_{2}=0}
c
1
=
−
c
2
e
t
=
0
{\displaystyle c_{1}=-c_{2}\mathrm {e} ^{t}=0}
Kahden funktion lineaarisen riippumattomuuden tarkistaminen tai todistaminen esimerkin 2 tapaan on kuitenkin varsin työlästä. Esitetään seuraavaksi menetelmä, jonka avulla voidaan nopeammin todeta lineaarinen riippuvuus tai riippumattomuus.
Määritelmä 5: Wronskin determinantti
muokkaa
Olkoon
x
1
,
x
2
∈
C
1
(
]
a
,
b
[
)
{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{1}(]a,b[)}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Wronskin determinantti
W
(
x
1
,
x
2
)
:
]
a
,
b
[
→
R
{\displaystyle W(x_{1},x_{2}):\,]a,b[\,\to \mathbb {R} }
[3]
W
(
x
1
,
x
2
)
(
t
)
=
det
[
x
1
(
t
)
x
1
′
(
t
)
x
2
(
t
)
x
2
′
(
t
)
]
=
x
1
(
t
)
x
2
′
(
t
)
−
x
2
(
t
)
x
1
′
(
t
)
{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{1}'(t)\\x_{2}(t)&x_{2}'(t)\end{matrix}}\right]\\&=x_{1}(t)x_{2}'(t)-x_{2}(t)x_{1}'(t)\end{aligned}}}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Jos
x
1
,
x
2
∈
C
1
(
]
a
,
b
[
)
{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{1}(]a,b[)}
on olemassa luku
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Lemman 1 (epäsuora) todistus
Antiteesi: On olemassa luku
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
määritelmän 4 mukaisesti, on olemassa vakio
c
∈
R
{\displaystyle c\in \mathbb {R} }
x
1
(
t
)
=
c
x
2
(
t
)
{\displaystyle x_{1}(t)=cx_{2}(t)}
x
2
(
t
)
=
c
x
1
(
t
)
{\displaystyle x_{2}(t)=cx_{1}(t)}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
2
(
t
)
=
c
x
1
(
t
)
{\displaystyle x_{2}(t)=cx_{1}(t)}
x
2
′
(
t
)
=
c
x
1
′
(
t
)
{\displaystyle x_{2}'(t)=cx_{1}'(t)}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
millä tahansa luvulla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
)
=
x
1
(
t
)
x
2
′
(
t
)
−
x
1
′
(
t
)
x
2
(
t
)
=
c
x
1
(
t
)
x
1
′
(
t
)
−
c
x
1
(
t
)
x
1
′
(
t
)
=
0
{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=x_{1}(t)x_{2}'(t)-x_{1}'(t)x_{2}(t)\\&=cx_{1}(t)x_{1}'(t)-cx_{1}(t)x_{1}'(t)\\&=0\end{aligned}}}
Ts. ei ole olemassa lukua
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
Q.E.D.
HUOM! Lemma 1 ei toimi toisin päin. Ts. jos
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
(
x
1
,
x
2
)
(
t
)
≠
0
{\displaystyle W(x_{1},x_{2})(t)\neq 0}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
1
,
x
2
:
R
→
R
{\displaystyle x_{1},x_{2}:\mathbb {R} \to \mathbb {R} }
x
1
(
t
)
=
{
t
2
,
jos
t
≥
0
0
,
jos
t
<
0
{\displaystyle x_{1}(t)={\begin{cases}t^{2},&{\text{jos }}t\geq 0\\0,&{\text{jos }}t<0\end{cases}}}
ja
x
2
(
t
)
=
{
0
,
jos
t
≥
0
t
2
,
jos
t
<
0
,
{\displaystyle x_{2}(t)={\begin{cases}0,&{\text{jos }}t\geq 0\\t^{2},&{\text{jos }}t<0,\end{cases}}}
niin
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
(
x
1
,
x
2
)
(
t
)
=
0
{\displaystyle W(x_{1},x_{2})(t)=0}
t
∈
R
{\displaystyle t\in \mathbb {R} }
Olkoon funktiot
x
1
,
x
2
∈
C
2
(
]
a
,
b
[
)
{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{2}(]a,b[)}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
Lauseen 2 todistus
''
⇒
{\displaystyle \Rightarrow }
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
x
2
(
t
0
)
≠
0
{\displaystyle x_{2}(t_{0})\neq 0}
x
2
≡
0
{\displaystyle x_{2}\equiv 0}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
2
{\displaystyle x_{2}}
δ
>
0
{\displaystyle \delta >0}
x
2
(
t
)
≠
0
{\displaystyle x_{2}(t)\neq 0}
kaikilla
t
∈
]
t
0
−
δ
,
t
0
+
δ
[
{\displaystyle t\in \,]t_{0}-\delta ,t_{0}+\delta [}
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
Antiteesi:
W
(
x
1
,
x
2
)
(
t
0
)
=
0
{\displaystyle W(x_{1},x_{2})(t_{0})=0}
W
(
x
1
,
x
2
)
(
t
)
=
0
{\displaystyle W(x_{1},x_{2})(t)=0}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
d
d
t
(
x
1
x
2
)
(
t
)
=
x
1
′
(
t
)
x
2
(
t
)
−
x
2
′
(
t
)
x
1
(
t
)
x
2
(
t
)
2
=
−
W
(
x
1
,
x
2
)
(
t
)
x
2
(
t
)
2
=
0
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {x_{1}}{x_{2}}}\right)(t)={\frac {x_{1}'(t)x_{2}(t)-x_{2}'(t)x_{1}(t)}{x_{2}(t)^{2}}}=-{\frac {W(x_{1},x_{2})(t)}{x_{2}(t)^{2}}}=0}
kaikilla
t
∈
]
t
0
−
δ
,
t
0
+
δ
[
{\displaystyle t\in \,]t_{0}-\delta ,t_{0}+\delta [}
.
Tällöin
(
x
1
/
x
2
)
(
t
)
=
c
=
vakio
{\displaystyle (x_{1}/x_{2})(t)=c={\text{vakio}}}
x
1
(
t
)
=
c
x
2
(
t
)
{\displaystyle x_{1}(t)=cx_{2}(t)}
t
∈
]
t
0
−
δ
,
t
0
+
δ
[
{\displaystyle t\in \,]t_{0}-\delta ,t_{0}+\delta [}
{
x
1
(
t
0
)
=
c
x
2
(
t
)
x
1
′
(
t
0
)
=
c
x
2
′
(
t
)
{\displaystyle {\begin{cases}x_{1}(t_{0})=cx_{2}(t)\\x_{1}'(t_{0})=cx_{2}'(t)\end{cases}}}
(Tämä pätee, kunhan
δ
{\displaystyle \delta }
x
1
{\displaystyle x_{1}}
c
x
2
{\displaystyle cx_{2}}
[4]
x
1
(
t
)
=
c
x
2
(
t
)
{\displaystyle x_{1}(t)=cx_{2}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Tämä on määritelmän 4 mukaan ristiriita.
''
⇐
{\displaystyle \Leftarrow }
Lemma 1
Q.E.D.
Seuraavat ehdot ovat yhtäpitäviä[5]
x
1
,
x
2
∈
C
2
(
]
a
,
b
[
)
{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{2}(]a,b[)}
Funktiot
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
ratkaisukannan .
Funktiot
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
lineaarisesti riippumattomia .
W
(
x
1
,
x
2
)
(
t
0
)
≠
0
{\displaystyle W(x_{1},x_{2})(t_{0})\neq 0}
jollakin
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
W
(
x
1
,
x
2
)
(
t
)
≠
0
{\displaystyle W(x_{1},x_{2})(t)\neq 0}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Lauseen 3 osatodistus
1.
⇒
{\displaystyle \Rightarrow }
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Antiteesi: Funktiot
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
c
∈
R
{\displaystyle c\in \mathbb {R} }
x
1
(
t
)
=
c
x
2
(
t
)
{\displaystyle x_{1}(t)=cx_{2}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
=
c
1
c
x
2
(
t
)
+
c
2
x
2
(
t
)
=
(
c
1
c
+
c
2
)
x
2
(
t
)
=:
c
3
x
2
(
t
)
,
(
♢
)
{\displaystyle {\begin{aligned}x(t)&=c_{1}x_{1}(t)+c_{2}x_{2}(t)\\&=c_{1}cx_{2}(t)+c_{2}x_{2}(t)\\&=(c_{1}c+c_{2})x_{2}(t)\\&=:c_{3}x_{2}(t),\qquad (\diamondsuit )\end{aligned}}}
missä
c
3
∈
R
{\displaystyle c_{3}\in \mathbb {R} }
t
0
∈
]
a
,
b
[
{\displaystyle t_{0}\in \,]a,b[}
x
2
(
t
0
)
≠
0
{\displaystyle x_{2}(t_{0})\neq 0}
{
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
x
(
t
0
)
=
0
x
′
(
t
0
)
=
1.
{\displaystyle {\begin{cases}x''(t)+p(t)x'(t)+r(t)x(t)=0\\x(t_{0})=0\\x'(t_{0})=1.\end{cases}}}
Yhtälön (
♢
{\displaystyle \diamondsuit }
x
(
t
)
=
c
3
x
2
(
t
)
{\displaystyle x(t)=c_{3}x_{2}(t)}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
c
3
x
2
(
t
0
)
=
x
(
t
0
)
=
0
{\displaystyle c_{3}x_{2}(t_{0})=x(t_{0})=0}
c
3
=
0
{\displaystyle c_{3}=0}
x
(
t
)
≡
0
{\displaystyle x(t)\equiv 0}
x
′
(
t
0
)
=
0
≠
1
{\displaystyle x'(t_{0})=0\neq 1}
x
2
(
t
0
)
=
0
{\displaystyle x_{2}(t_{0})=0}
{
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
x
(
t
0
)
=
1
x
′
(
t
0
)
=
1.
{\displaystyle {\begin{cases}x''(t)+p(t)x'(t)+r(t)x(t)=0\\x(t_{0})=1\\x'(t_{0})=1.\end{cases}}}
Yhtälön (
♢
{\displaystyle \diamondsuit }
x
(
t
)
=
c
3
x
2
(
t
)
{\displaystyle x(t)=c_{3}x_{2}(t)}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
0
=
c
3
x
2
(
t
0
)
=
x
(
t
0
)
=
1
{\displaystyle 0=c_{3}x_{2}(t_{0})=x(t_{0})=1}
Antiteesi johtaa ristiriitaan, joten väite 2 seuraa väitteestä 1.
2.
⇒
{\displaystyle \Rightarrow }
Lause 2
3.
⇒
{\displaystyle \Rightarrow }
4.
⇒
{\displaystyle \Rightarrow }
W
(
x
1
,
x
2
)
(
t
)
≠
0
{\displaystyle W(x_{1},x_{2})(t)\neq 0}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
′
(
t
)
+
p
(
t
)
W
(
t
)
=
0
{\displaystyle W'(t)+p(t)W(t)=0}
Näin ollen[6]
W
(
t
)
=
W
(
t
0
)
e
−
∫
t
0
t
p
(
s
)
d
s
=
C
e
−
∫
t
0
t
p
(
s
)
d
s
⏟
=
μ
(
t
)
=
C
μ
(
t
)
.
{\displaystyle {\begin{aligned}W(t)&=W(t_{0})\mathrm {e} ^{-\int _{t_{0}}^{t}p(s)\,\mathrm {d} s}\\&=C\underbrace {\mathrm {e} ^{-\int _{t_{0}}^{t}p(s)\,\mathrm {d} s}} _{=\mu (t)}\\&=C\mu (t).\end{aligned}}}
Olkoon
x
{\displaystyle x}
W
(
x
1
,
x
2
)
(
t
)
=
C
μ
(
t
)
{\displaystyle W(x_{1},x_{2})(t)=C\mu (t)}
C
≠
0
{\displaystyle C\neq 0}
W
(
x
,
x
1
)
(
t
)
=
C
1
μ
(
t
)
{\displaystyle W(x,x_{1})(t)=C_{1}\mu (t)}
W
(
x
,
x
2
)
(
t
)
=
C
2
μ
(
t
)
{\displaystyle W(x,x_{2})(t)=C_{2}\mu (t)}
C
1
,
C
2
≠
0
{\displaystyle C_{1},C_{2}\neq 0}
(
C
1
x
2
(
t
)
−
C
2
x
1
(
t
)
)
μ
(
t
)
=
W
(
x
,
x
1
)
(
t
)
⋅
x
2
(
t
)
−
W
(
x
,
x
2
)
(
t
)
⋅
x
1
(
t
)
=
[
x
(
t
)
x
1
′
(
t
)
−
x
′
(
t
)
x
1
(
t
)
]
x
2
(
t
)
−
[
x
(
t
)
x
2
′
(
t
)
−
x
′
(
t
)
x
2
(
t
)
]
x
1
(
t
)
=
x
(
t
)
x
1
′
(
t
)
x
2
(
t
)
−
x
(
t
)
x
2
′
(
t
)
x
1
(
t
)
−
x
′
(
t
)
x
1
(
t
)
x
2
(
t
)
+
x
′
(
t
)
x
2
(
t
)
x
1
(
t
)
=
−
x
(
t
)
[
x
2
′
(
t
)
x
1
(
t
)
−
x
1
′
(
t
)
x
2
(
t
)
]
=
−
x
(
t
)
W
(
x
1
,
x
2
)
(
t
)
=
−
C
x
(
t
)
μ
(
t
)
{\displaystyle {\begin{aligned}(C_{1}x_{2}(t)-C_{2}x_{1}(t))\mu (t)&=W(x,x_{1})(t)\cdot x_{2}(t)-W(x,x_{2})(t)\cdot x_{1}(t)\\&=[x(t)x_{1}'(t)-x'(t)x_{1}(t)]x_{2}(t)-[x(t)x_{2}'(t)-x'(t)x_{2}(t)]x_{1}(t)\\&=x(t)x_{1}'(t)x_{2}(t)-x(t)x_{2}'(t)x_{1}(t)-x'(t)x_{1}(t)x_{2}(t)+x'(t)x_{2}(t)x_{1}(t)\\&=-x(t)[x_{2}'(t)x_{1}(t)-x_{1}'(t)x_{2}(t)]\\&=-x(t)W(x_{1},x_{2})(t)\\&=-Cx(t)\mu (t)\end{aligned}}}
Koska
C
≠
0
{\displaystyle C\neq 0}
μ
(
t
)
>
0
{\displaystyle \mu (t)>0}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
(
t
)
=
C
2
C
x
1
(
t
)
−
C
1
C
x
2
(
t
)
.
{\displaystyle x(t)={\frac {C_{2}}{C}}x_{1}(t)-{\frac {C_{1}}{C}}x_{2}(t).}
Määritelmän 3 nojalla funktiot
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Q.E.D.
Ratkaistaan differentiaaliyhtälö
x
″
(
t
)
+
x
(
t
)
=
0
{\displaystyle x''(t)+x(t)=0}
x
1
(
t
)
=
sin
t
{\displaystyle x_{1}(t)=\sin t}
x
1
(
t
)
=
cos
t
{\displaystyle x_{1}(t)=\cos t}
W
(
x
1
,
x
2
)
(
t
)
=
−
sin
2
t
−
cos
2
t
≡
−
1
≠
0
{\displaystyle W(x_{1},x_{2})(t)=-\sin ^{2}t-\cos ^{2}t\equiv -1\neq 0}
kaikilla
t
∈
R
{\displaystyle t\in \mathbb {R} }
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
(
t
)
=
c
1
sin
t
+
c
2
cos
t
{\displaystyle x(t)=c_{1}\sin t+c_{2}\cos t}
missä
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
Ratkaistaan differentiaaliyhtälö
x
″
(
t
)
−
x
(
t
)
=
0
{\displaystyle x''(t)-x(t)=0}
x
1
(
t
)
=
e
t
{\displaystyle x_{1}(t)=\mathrm {e} ^{t}}
x
1
(
t
)
=
e
−
t
{\displaystyle x_{1}(t)=\mathrm {e} ^{-t}}
W
(
x
1
,
x
2
)
(
t
)
=
e
t
⋅
(
−
e
−
t
)
−
e
−
t
⋅
e
t
=
−
2
≠
0
{\displaystyle W(x_{1},x_{2})(t)=\mathrm {e} ^{t}\cdot (-\mathrm {e} ^{-t})-\mathrm {e} ^{-t}\cdot \mathrm {e} ^{t}=-2\neq 0}
Tällöin
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
(
t
)
=
c
1
e
t
+
c
2
e
−
t
{\displaystyle x(t)=c_{1}\mathrm {e} ^{t}+c_{2}\mathrm {e} ^{-t}}
missä
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
Olkoon
p
,
r
:
]
a
,
b
[
→
R
{\displaystyle p,r:\,]a,b[\,\to \mathbb {R} }
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
on aina olemassa ratkaisukanta.
Ratkaisukannan löytäminen kertaluvun pudotuksella
muokkaa
Edellisen luvun perusteella homogeenisen differentiaaliyhtälön
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
kaikki ratkaisut löytyvät, jos löydetään kaksi toisistaan lineaarisesti riippumatonta ratkaisua. Tässä luvussa todetaan, että riittää löytää vain yksi ei-triviaali ratkaisu.[1]
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
1
{\displaystyle x_{1}}
x
1
(
t
)
≠
0
{\displaystyle x_{1}(t)\neq 0}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
2
(
t
)
=
v
(
t
)
x
1
(
t
)
{\displaystyle x_{2}(t)=v(t)x_{1}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
jollekin funktiolle
v
:
]
a
,
b
[
→
R
{\displaystyle v:\,]a,b[\,\to \mathbb {R} }
v
{\displaystyle v}
0
=
x
2
″
(
t
)
+
p
(
t
)
x
2
′
(
t
)
+
r
(
t
)
x
2
(
t
)
=
v
″
(
t
)
x
1
(
t
)
+
2
v
′
(
t
)
x
1
′
(
t
)
+
v
(
t
)
x
1
″
(
t
)
+
p
(
t
)
[
v
′
(
t
)
x
1
(
t
)
+
v
(
t
)
x
1
′
(
t
)
]
+
r
(
t
)
v
(
t
)
x
1
(
t
)
=
v
(
t
)
[
x
1
″
(
t
)
+
p
(
t
)
x
1
′
+
r
(
t
)
x
1
(
t
)
]
⏟
=
0
+
v
″
(
t
)
x
1
(
t
)
+
2
v
′
(
t
)
x
1
′
(
t
)
+
p
(
t
)
x
1
(
t
)
=
v
″
(
t
)
x
1
(
t
)
+
2
v
′
(
t
)
x
1
′
(
t
)
+
p
(
t
)
x
1
(
t
)
{\displaystyle {\begin{aligned}0&=x_{2}''(t)+p(t)x_{2}'(t)+r(t)x_{2}(t)\\&=v''(t)x_{1}(t)+2v'(t)x_{1}'(t)+v(t)x_{1}''(t)+p(t)[v'(t)x_{1}(t)+v(t)x_{1}'(t)]+r(t)v(t)x_{1}(t)\\&=v(t)\underbrace {[x_{1}''(t)+p(t)x_{1}'+r(t)x_{1}(t)]} _{=0}+v''(t)x_{1}(t)+2v'(t)x_{1}'(t)+p(t)x_{1}(t)\\&=v''(t)x_{1}(t)+2v'(t)x'_{1}(t)+p(t)x_{1}(t)\end{aligned}}}
Jaetaan yhtälö puolittain
x
1
{\displaystyle x_{1}}
x
1
(
t
)
≠
0
{\displaystyle x_{1}(t)\neq 0}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
s
:
]
a
,
b
[
→
R
{\displaystyle s:\,]a,b[\,\to \mathbb {R} }
s
(
t
)
=
2
x
1
′
(
t
)
x
1
(
t
)
+
p
(
t
)
{\displaystyle s(t)=2{\frac {x_{1}'(t)}{x_{1}(t)}}+p(t)}
Tällöin
v
″
(
t
)
+
2
v
′
(
t
)
x
1
′
(
t
)
x
1
(
t
)
+
p
(
t
)
v
′
(
t
)
=
0
v
″
(
t
)
+
(
2
x
1
′
(
t
)
x
1
(
t
)
+
p
(
t
)
)
⏟
=
s
(
t
)
v
′
(
t
)
=
0
v
″
(
t
)
+
s
(
t
)
v
′
(
t
)
=
0
{\displaystyle {\begin{aligned}v''(t)+{\frac {2v'(t)x_{1}'(t)}{x_{1}(t)}}+p(t)v'(t)&=0\\v''(t)+\underbrace {\left(2{\frac {x_{1}'(t)}{x_{1}(t)}}+p(t)\right)} _{=s(t)}v'(t)&=0\\v''(t)+s(t)v'(t)&=0\end{aligned}}}
Merkitään edelleen
z
(
t
)
=
v
′
(
t
)
{\displaystyle z(t)=v'(t)}
z
′
(
t
)
+
s
(
t
)
z
(
t
)
=
0
{\displaystyle z'(t)+s(t)z(t)=0}
Tiedetään[7]
z
(
t
)
=
e
−
∫
t
0
t
s
(
r
)
d
r
{\displaystyle z(t)=\mathrm {e} ^{-\int _{t_{0}}^{t}s(r)\,\mathrm {d} r}}
Siispä
v
(
t
)
=
∫
t
0
t
z
(
u
)
d
u
=
∫
t
0
t
e
−
∫
u
0
u
s
(
r
)
d
r
d
u
{\displaystyle v(t)=\int _{t_{0}}^{t}z(u)\,\mathrm {d} u=\int _{t_{0}}^{t}\mathrm {e} ^{-\int _{u_{0}}^{u}s(r)\,\mathrm {d} r}\,\mathrm {d} u}
Tällöin
x
2
(
t
)
=
v
(
t
)
x
1
(
t
)
{\displaystyle x_{2}(t)=v(t)x_{1}(t)}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
(
x
1
,
x
2
)
(
t
)
=
det
[
x
1
(
t
)
x
2
(
t
)
x
1
′
(
t
)
x
2
′
(
t
)
]
=
det
[
x
1
(
t
)
v
(
t
)
x
1
(
t
)
x
1
′
(
t
)
v
′
(
t
)
x
1
(
t
)
+
v
(
t
)
x
1
′
(
t
)
]
=
x
1
(
t
)
[
v
′
(
t
)
x
1
(
t
)
+
v
(
t
)
x
1
′
(
t
)
]
−
x
1
′
(
t
)
v
(
t
)
x
1
(
t
)
=
v
′
(
t
)
x
1
2
(
t
)
=
z
(
t
)
x
1
2
(
t
)
=
e
−
∫
t
0
t
s
(
t
)
d
s
⋅
x
1
2
(
t
)
≠
0
{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{2}(t)\\x_{1}'(t)&x_{2}'(t)\end{matrix}}\right]\\&=\det \left[{\begin{matrix}x_{1}(t)&v(t)x_{1}(t)\\x_{1}'(t)&v'(t)x_{1}(t)+v(t)x_{1}'(t)\end{matrix}}\right]\\&=x_{1}(t)[v'(t)x_{1}(t)+v(t)x_{1}'(t)]-x_{1}'(t)v(t)x_{1}(t)\\&=v'(t)x_{1}^{2}(t)\\&=z(t)x_{1}^{2}(t)\\&=\mathrm {e} ^{-\int _{t_{0}}^{t}s(t)\,\mathrm {d} s}\cdot x_{1}^{2}(t)\neq 0\end{aligned}}}
Ratkaistaan differentiaaliyhtälö
x
″
(
t
)
−
1
t
x
′
(
t
)
+
1
t
2
x
(
t
)
=
0
,
t
>
0
{\displaystyle x''(t)-{\frac {1}{t}}x'(t)+{\frac {1}{t^{2}}}x(t)=0,\quad t>0}
Havaitaan, että eräs ratkaisu on
x
1
(
t
)
=
t
{\displaystyle x_{1}(t)=t}
x
1
′
(
t
)
=
1
{\displaystyle x_{1}'(t)=1}
x
1
″
(
t
)
=
0
{\displaystyle x_{1}''(t)=0}
0
−
1
t
⋅
1
+
1
t
2
⋅
t
=
−
1
t
+
1
t
=
0
{\displaystyle 0-{\frac {1}{t}}\cdot 1+{\frac {1}{t^{2}}}\cdot t=-{\frac {1}{t}}+{\frac {1}{t}}=0}
Olkoon
x
2
(
t
)
=
v
(
t
)
x
1
(
t
)
=
v
(
t
)
t
{\displaystyle x_{2}(t)=v(t)x_{1}(t)=v(t)t}
x
2
′
(
t
)
=
v
′
(
t
)
t
+
v
(
t
)
{\displaystyle x_{2}'(t)=v'(t)t+v(t)}
x
2
″
(
t
)
=
v
″
(
t
)
t
+
2
v
′
(
t
)
{\displaystyle x_{2}''(t)=v''(t)t+2v'(t)}
0
=
x
2
″
(
t
)
−
1
t
x
2
′
(
t
)
+
1
t
2
x
2
(
t
)
=
v
″
(
t
)
t
+
2
v
′
(
t
)
−
1
t
[
v
′
(
t
)
t
+
v
(
t
)
]
+
1
t
2
v
(
t
)
t
=
t
v
″
(
t
)
+
2
v
′
(
t
)
−
v
′
(
t
)
−
v
(
t
)
t
+
v
(
t
)
t
=
t
v
″
(
t
)
+
v
′
(
t
)
{\displaystyle {\begin{aligned}0&=x_{2}''(t)-{\frac {1}{t}}x_{2}'(t)+{\frac {1}{t^{2}}}x_{2}(t)\\&=v''(t)t+2v'(t)-{\frac {1}{t}}[v'(t)t+v(t)]+{\frac {1}{t^{2}}}v(t)t\\&=tv''(t)+2v'(t)-v'(t)-{\frac {v(t)}{t}}+{\frac {v(t)}{t}}\\&=tv''(t)+v'(t)\end{aligned}}}
Merkitään
z
(
t
)
=
v
′
(
t
)
{\displaystyle z(t)=v'(t)}
z
′
(
t
)
+
1
t
z
(
t
)
=
0
z
(
t
)
=
e
−
∫
1
t
d
t
=
e
−
ln
t
+
C
1
=
e
C
1
e
ln
t
=
C
1
t
{\displaystyle {\begin{aligned}z'(t)+{\frac {1}{t}}z(t)&=0\\z(t)&=\mathrm {e} ^{-\int {\frac {1}{t}}\,\mathrm {d} t}\\&=\mathrm {e} ^{-\ln t+C_{1}}={\frac {\mathrm {e} ^{C_{1}}}{\mathrm {e} ^{\ln t}}}\\&={\frac {C_{1}}{t}}\end{aligned}}}
Tällöin
v
(
t
)
=
∫
C
1
t
d
t
=
C
1
ln
t
+
C
2
{\displaystyle v(t)=\int {\frac {C_{1}}{t}}\,\mathrm {d} t=C_{1}\ln t+C_{2}}
missä
C
1
,
C
2
∈
R
{\displaystyle C_{1},C_{2}\in \mathbb {R} }
x
2
(
t
)
=
v
(
t
)
t
=
(
C
1
ln
t
+
C
2
)
t
=
C
1
t
ln
t
+
C
2
t
{\displaystyle x_{2}(t)=v(t)t=(C_{1}\ln t+C_{2})t=C_{1}t\ln t+C_{2}t}
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
=
c
1
t
+
c
2
(
C
1
t
ln
t
+
C
2
t
)
{\displaystyle {\begin{aligned}x(t)&=c_{1}x_{1}(t)+c_{2}x_{2}(t)\\&=c_{1}t+c_{2}(C_{1}t\ln t+C_{2}t)\end{aligned}}}
Muotoillaan vakioita uudelleen siten, että
c
1
+
c
2
C
2
=
A
{\displaystyle c_{1}+c_{2}C_{2}=A}
C
1
c
2
=
B
{\displaystyle C_{1}c_{2}=B}
x
:
]
0
,
∞
[
→
R
{\displaystyle x:\,]0,\infty [\,\to \mathbb {R} }
x
(
t
)
=
A
t
+
B
t
ln
t
{\displaystyle x(t)=At+Bt\ln t}
Ratkaistaan 1. asteen Legendren differentiaaliyhtälö (
λ
=
1
{\displaystyle \lambda =1}
(
1
−
t
2
)
x
″
(
t
)
−
2
t
x
′
(
t
)
+
2
x
(
t
)
=
0
,
0
<
t
<
1
{\displaystyle (1-t^{2})x''(t)-2tx'(t)+2x(t)=0,\qquad 0<t<1}
Eräs ratkaisu on
x
1
(
t
)
=
t
{\displaystyle x_{1}(t)=t}
x
1
′
(
t
)
=
1
{\displaystyle x_{1}'(t)=1}
x
1
″
(
t
)
=
0
{\displaystyle x_{1}''(t)=0}
(
1
−
t
2
)
⋅
0
−
2
t
⋅
1
+
2
t
=
−
2
t
+
2
t
=
0
{\displaystyle (1-t^{2})\cdot 0-2t\cdot 1+2t=-2t+2t=0}
Olkoon
x
2
(
t
)
=
v
(
t
)
x
1
(
t
)
=
v
(
t
)
t
{\displaystyle x_{2}(t)=v(t)x_{1}(t)=v(t)t}
0
=
(
1
−
t
2
)
x
2
″
(
t
)
−
2
t
x
2
′
(
t
)
+
2
x
2
(
t
)
=
(
1
−
t
2
)
[
v
″
(
t
)
t
+
2
v
′
(
t
)
]
−
2
t
[
v
′
(
t
)
+
v
(
t
)
]
+
2
v
(
t
)
t
=
(
1
−
t
2
)
t
v
″
(
t
)
+
2
(
1
−
t
2
)
v
′
(
t
)
−
2
t
2
v
′
(
t
)
{\displaystyle {\begin{aligned}0&=(1-t^{2})x_{2}''(t)-2tx_{2}'(t)+2x_{2}(t)\\&=(1-t^{2})[v''(t)t+2v'(t)]-2t[v'(t)+v(t)]+2v(t)t\\&=(1-t^{2})tv''(t)+2(1-t^{2})v'(t)-2t^{2}v'(t)\end{aligned}}}
Koska
0
<
t
<
1
{\displaystyle 0<t<1}
(
1
−
t
2
)
t
{\displaystyle (1-t^{2})t}
v
″
(
t
)
+
(
2
t
−
2
t
1
−
t
2
)
v
′
(
t
)
=
0
{\displaystyle {\begin{aligned}v''(t)+\left({\frac {2}{t}}-{\frac {2t}{1-t^{2}}}\right)v'(t)=0\end{aligned}}}
Merkitään
z
(
t
)
=
v
′
(
t
)
{\displaystyle z(t)=v'(t)}
z
′
(
t
)
+
(
2
t
−
2
t
1
−
t
2
)
z
(
t
)
=
0
{\displaystyle z'(t)+\left({\frac {2}{t}}-{\frac {2t}{1-t^{2}}}\right)z(t)=0}
Tällöin
z
(
t
)
=
exp
[
−
∫
2
t
d
t
+
∫
2
t
1
−
t
2
d
t
]
=
exp
[
−
ln
t
2
−
ln
(
1
−
t
2
)
+
C
1
]
=
exp
(
C
1
)
⋅
exp
[
−
ln
(
t
2
(
1
−
t
2
)
)
]
=
C
~
1
t
2
(
1
−
t
2
)
,
{\displaystyle {\begin{aligned}z(t)&=\exp \left[-\int {\frac {2}{t}}\,\mathrm {d} t+\int {\frac {2t}{1-t^{2}}}\,\mathrm {d} t\right]\\&=\exp \left[-\ln t^{2}-\ln(1-t^{2})+C_{1}\right]\\&=\exp(C_{1})\cdot \exp \left[-\ln \left(t^{2}(1-t^{2})\right)\right]\\&={\frac {{\tilde {C}}_{1}}{t^{2}(1-t^{2})}},\end{aligned}}}
missä
C
~
1
∈
R
{\displaystyle {\tilde {C}}_{1}\in \mathbb {R} }
z
{\displaystyle z}
osamurtokehitelmä :
z
(
t
)
=
C
~
1
t
2
(
1
−
t
2
)
=
C
~
1
(
1
t
2
+
1
2
(
1
−
t
)
+
1
2
(
1
+
t
)
)
{\displaystyle z(t)={\frac {{\tilde {C}}_{1}}{t^{2}(1-t^{2})}}={\tilde {C}}_{1}\left({\frac {1}{t^{2}}}+{\frac {1}{2(1-t)}}+{\frac {1}{2(1+t)}}\right)}
Ratkaistaan
v
{\displaystyle v}
v
(
t
)
=
∫
z
(
t
)
d
t
=
∫
C
~
1
t
2
(
1
−
t
2
)
d
t
=
C
~
1
(
∫
d
t
t
2
+
1
2
∫
d
t
1
−
t
+
1
2
∫
d
t
1
+
t
)
=
C
~
1
(
−
1
t
−
1
2
ln
(
1
−
t
)
+
1
2
ln
(
1
+
t
)
+
C
2
)
=
C
~
1
(
−
1
t
+
1
2
ln
(
1
+
t
1
−
t
)
+
C
2
)
,
{\displaystyle {\begin{aligned}v(t)&=\int z(t)\,\mathrm {d} t=\int {\frac {{\tilde {C}}_{1}}{t^{2}(1-t^{2})}}\,\mathrm {d} t\\&={\tilde {C}}_{1}\left(\int {\frac {\mathrm {d} t}{t^{2}}}+{\frac {1}{2}}\int {\frac {\mathrm {d} t}{1-t}}+{\frac {1}{2}}\int {\frac {\mathrm {d} t}{1+t}}\right)\\&={\tilde {C}}_{1}\left(-{\frac {1}{t}}-{\frac {1}{2}}\ln(1-t)+{\frac {1}{2}}\ln(1+t)+C_{2}\right)\\&={\tilde {C}}_{1}\left(-{\frac {1}{t}}+{\frac {1}{2}}\ln \left({\frac {1+t}{1-t}}\right)+C_{2}\right),\end{aligned}}}
missä
C
2
∈
R
{\displaystyle C_{2}\in \mathbb {R} }
x
2
(
t
)
=
v
(
t
)
t
=
C
~
1
(
−
1
+
t
2
ln
(
1
+
t
1
−
t
)
+
C
2
t
)
{\displaystyle x_{2}(t)=v(t)t={\tilde {C}}_{1}\left(-1+{\frac {t}{2}}\ln \left({\frac {1+t}{1-t}}\right)+C_{2}t\right)}
Yhdistetään ratkaisut:
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
=
c
1
t
+
c
2
C
~
1
(
t
2
ln
(
1
+
t
1
−
t
)
+
C
2
t
−
1
)
{\displaystyle {\begin{aligned}x(t)&=c_{1}x_{1}(t)+c_{2}x_{2}(t)\\&=c_{1}t+c_{2}{\tilde {C}}_{1}\left({\frac {t}{2}}\ln \left({\frac {1+t}{1-t}}\right)+C_{2}t-1\right)\end{aligned}}}
Muotoillaan vakioita uudelleen siten, että
c
1
+
c
2
C
~
1
C
2
=
A
{\displaystyle c_{1}+c_{2}{\tilde {C}}_{1}C_{2}=A}
c
2
C
~
1
=
B
{\displaystyle c_{2}{\tilde {C}}_{1}=B}
x
:
]
0
,
1
[
→
R
{\displaystyle x:\,]0,1[\,\to \mathbb {R} }
x
(
t
)
=
A
t
+
B
(
t
2
ln
(
1
+
t
1
−
t
−
1
)
)
{\displaystyle x(t)=At+B\left({\frac {t}{2}}\ln \left({\frac {1+t}{1-t}}-1\right)\right)}
Vakiokertoiminen homogeeninen differentiaaliyhtälö
muokkaa
Tarkastellaan vakiokertoimista homogeenista differentiaaliyhtälöä
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
0
{\displaystyle x''(t)+bx'(t)+cx(t)=0}
missä
b
,
c
∈
R
{\displaystyle b,c\in \mathbb {R} }
x
(
t
)
=
e
λ
t
{\displaystyle x(t)=\mathrm {e} ^{\lambda t}}
λ
∈
R
{\displaystyle \lambda \in \mathbb {R} }
x
′
(
t
)
=
λ
e
λ
t
=
λ
x
(
t
)
x
″
(
t
)
=
λ
2
e
λ
t
=
λ
2
x
(
t
)
{\displaystyle {\begin{aligned}x'(t)&=\lambda \mathrm {e} ^{\lambda t}=\lambda x(t)\\x''(t)&=\lambda ^{2}\mathrm {e} ^{\lambda t}=\lambda ^{2}x(t)\end{aligned}}}
Sijoitetaan nämä DY:öön:
λ
2
x
(
t
)
+
b
λ
x
(
t
)
+
c
x
(
t
)
=
0
|
|
:
x
(
t
)
(
≠
0
∀
t
∈
R
)
λ
2
+
b
λ
+
c
=
0
(
♡
)
{\displaystyle {\begin{aligned}\lambda ^{2}x(t)+b\lambda x(t)+cx(t)&=0\quad ||:x(t)~(\neq 0~\forall ~t\in \mathbb {R} )\\\lambda ^{2}+b\lambda +c&=0\quad (\heartsuit )\end{aligned}}}
Siis jos luku
λ
{\displaystyle \lambda }
♡
{\displaystyle \heartsuit }
x
{\displaystyle x}
♡
{\displaystyle \heartsuit }
karakteristiseksi yhtälöksi , jonka ratkaisut ovat[8]
λ
1
=
−
b
+
b
2
−
4
c
2
{\displaystyle \lambda _{1}={\frac {-b+{\sqrt {b^{2}-4c}}}{2}}}
λ
2
=
−
b
−
b
2
−
4
c
2
{\displaystyle \lambda _{2}={\frac {-b-{\sqrt {b^{2}-4c}}}{2}}}
Jos
b
2
−
4
c
>
0
{\displaystyle b^{2}-4c>0}
muokkaa
Jos
b
2
−
4
c
>
0
{\displaystyle b^{2}-4c>0}
{
x
1
(
t
)
=
e
λ
1
t
x
2
(
t
)
=
e
λ
2
t
{\displaystyle {\begin{cases}x_{1}(t)=\mathrm {e} ^{\lambda _{1}t}\\x_{2}(t)=\mathrm {e} ^{\lambda _{2}t}\end{cases}}}
Lisäksi
W
(
x
1
,
x
2
)
(
t
)
=
det
[
e
λ
1
t
λ
1
e
λ
1
t
e
λ
2
t
λ
2
e
λ
2
t
]
=
(
λ
2
−
λ
1
)
e
(
λ
1
+
λ
2
)
t
≠
0
{\displaystyle W(x_{1},x_{2})(t)=\det \left[{\begin{matrix}\mathrm {e} ^{\lambda _{1}t}&\lambda _{1}\mathrm {e} ^{\lambda _{1}t}\\\mathrm {e} ^{\lambda _{2}t}&\lambda _{2}\mathrm {e} ^{\lambda _{2}t}\end{matrix}}\right]=(\lambda _{2}-\lambda _{1})\mathrm {e} ^{(\lambda _{1}+\lambda _{2})t}\neq 0}
sillä
λ
1
≠
λ
2
{\displaystyle \lambda _{1}\neq \lambda _{2}}
Lauseen 3 nojalla kaikki DY:n ratkaisut ovat muotoa
x
:
R
→
R
{\displaystyle x:\mathbb {R} \to \mathbb {R} }
x
(
t
)
=
c
1
e
λ
1
t
+
c
2
e
λ
2
t
{\displaystyle x(t)=c_{1}\mathrm {e} ^{\lambda _{1}t}+c_{2}\mathrm {e} ^{\lambda _{2}t}}
Jos
b
2
−
4
c
=
0
{\displaystyle b^{2}-4c=0}
muokkaa
Jos
b
2
−
4
c
=
0
{\displaystyle b^{2}-4c=0}
x
1
(
t
)
=
e
λ
t
=
e
−
b
t
/
2
{\displaystyle x_{1}(t)=\mathrm {e} ^{\lambda t}=\mathrm {e} ^{-bt/2}}
kertaluvun pudotuksella :
x
2
(
t
)
=
v
(
t
)
x
1
(
t
)
=
v
(
t
)
e
λ
t
{\displaystyle x_{2}(t)=v(t)x_{1}(t)=v(t)\mathrm {e} ^{\lambda t}}
v
″
(
t
)
+
(
2
x
1
′
(
t
)
x
1
(
t
)
+
b
)
v
′
(
t
)
=
0
v
″
(
t
)
+
(
2
λ
e
λ
t
e
λ
t
+
b
)
v
′
(
t
)
=
0
v
″
(
t
)
+
(
2
λ
+
b
)
v
′
(
t
)
=
0
|
λ
=
−
b
/
2
v
″
(
t
)
=
0
{\displaystyle {\begin{aligned}v''(t)+\left({\frac {2x_{1}'(t)}{x_{1}(t)}}+b\right)v'(t)&=0\\v''(t)+\left({\frac {2\lambda \mathrm {e} ^{\lambda t}}{\mathrm {e} ^{\lambda t}}}+b\right)v'(t)&=0\\v''(t)+(2\lambda +b)v'(t)&=0\qquad |~\lambda =-b/2\\v''(t)&=0\end{aligned}}}
Siis
v
(
t
)
=
C
1
t
+
C
2
{\displaystyle v(t)=C_{1}t+C_{2}}
C
1
,
C
2
∈
R
{\displaystyle C_{1},C_{2}\in \mathbb {R} }
C
1
=
1
{\displaystyle C_{1}=1}
C
2
=
0
{\displaystyle C_{2}=0}
x
:
R
→
R
{\displaystyle x:\mathbb {R} \to \mathbb {R} }
x
(
t
)
=
c
1
e
λ
t
+
c
2
t
e
λ
t
{\displaystyle x(t)=c_{1}\mathrm {e} ^{\lambda t}+c_{2}t\mathrm {e} ^{\lambda t}}
Jos
b
2
−
4
c
<
0
{\displaystyle b^{2}-4c<0}
muokkaa
Jos
b
2
−
4
c
<
0
{\displaystyle b^{2}-4c<0}
kompleksilukujen maailmaan. Karakteristisella yhtälöllä on kaksi kompleksista juurta:
{
λ
1
=
−
b
+
−
(
4
c
−
b
2
)
2
=
−
b
2
+
i
4
c
−
b
2
2
λ
2
=
−
b
−
−
(
4
c
−
b
2
)
2
=
−
b
2
−
i
4
c
−
b
2
2
{\displaystyle {\begin{cases}\displaystyle {\lambda _{1}={\frac {-b+{\sqrt {-(4c-b^{2})}}}{2}}=-{\frac {b}{2}}+\mathrm {i} {\frac {\sqrt {4c-b^{2}}}{2}}}\\\displaystyle {\lambda _{2}={\frac {-b-{\sqrt {-(4c-b^{2})}}}{2}}=-{\frac {b}{2}}-\mathrm {i} {\frac {\sqrt {4c-b^{2}}}{2}}}\end{cases}}}
Ratkaisut ovat siis
x
1
(
t
)
=
exp
(
−
b
2
t
+
i
4
c
−
b
2
2
t
)
=
e
−
b
t
/
2
⋅
e
i
4
c
−
b
2
2
t
{\displaystyle x_{1}(t)=\exp \left(-{\frac {b}{2}}t+\mathrm {i} {\frac {\sqrt {4c-b^{2}}}{2}}t\right)=\mathrm {e} ^{-bt/2}\cdot \mathrm {e} ^{\mathrm {i} {\frac {\sqrt {4c-b^{2}}}{2}}t}}
x
2
(
t
)
=
e
−
b
t
/
2
⋅
e
−
i
4
c
−
b
2
2
t
{\displaystyle x_{2}(t)=\mathrm {e} ^{-bt/2}\cdot \mathrm {e} ^{-\mathrm {i} {\frac {\sqrt {4c-b^{2}}}{2}}t}}
Lyhennyssyistä merkitään
α
=
−
b
2
{\displaystyle \alpha =-{\frac {b}{2}}}
β
=
4
c
−
b
2
2
{\displaystyle \beta ={\frac {\sqrt {4c-b^{2}}}{2}}}
4
c
−
b
2
>
0
{\displaystyle 4c-b^{2}>0}
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
x
1
(
t
)
=
e
α
t
⋅
e
i
β
t
{\displaystyle x_{1}(t)=\mathrm {e} ^{\alpha t}\cdot \mathrm {e} ^{\mathrm {i} \beta t}}
x
1
(
t
)
=
e
α
t
⋅
e
−
i
β
t
{\displaystyle x_{1}(t)=\mathrm {e} ^{\alpha t}\cdot \mathrm {e} ^{-\mathrm {i} \beta t}}
Yleinen ratkaisu voidaan kuitenkin sieventää siten, että se on puhtaasti reaalinen funktio. Eulerin lauseen mukaan
e
i
φ
=
cos
φ
+
i
sin
φ
{\displaystyle \mathrm {e} ^{\mathrm {i} \varphi }=\cos \varphi +\mathrm {i} \sin \varphi }
φ
{\displaystyle \varphi }
x
1
(
t
)
=
e
α
t
(
cos
(
β
t
)
+
i
sin
(
β
t
)
)
{\displaystyle x_{1}(t)=\mathrm {e} ^{\alpha t}\left(\cos(\beta t)+\mathrm {i} \sin(\beta t)\right)}
x
2
(
t
)
=
e
α
t
(
cos
(
−
β
t
)
+
i
sin
(
−
β
t
)
)
=
e
α
t
(
cos
(
β
t
)
−
i
sin
(
β
t
)
)
{\displaystyle x_{2}(t)=\mathrm {e} ^{\alpha t}\left(\cos(-\beta t)+\mathrm {i} \sin(-\beta t)\right)=\mathrm {e} ^{\alpha t}\left(\cos(\beta t)-\mathrm {i} \sin(\beta t)\right)}
[9]
Koska
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
lauseen 1 nojalla myös funktiot
x
1
+
x
2
{\displaystyle x_{1}+x_{2}}
x
1
−
x
2
{\displaystyle x_{1}-x_{2}}
(
x
1
+
x
2
)
(
t
)
=
e
α
t
(
cos
(
β
t
)
+
i
sin
(
β
t
)
)
+
e
α
t
(
cos
(
β
t
)
−
i
sin
(
β
t
)
)
=
2
e
α
t
cos
(
β
t
)
{\displaystyle {\begin{aligned}(x_{1}+x_{2})(t)&=\mathrm {e} ^{\alpha t}\left(\cos(\beta t)+\mathrm {i} \sin(\beta t)\right)+\mathrm {e} ^{\alpha t}\left(\cos(\beta t)-\mathrm {i} \sin(\beta t)\right)\\&=2\mathrm {e} ^{\alpha t}\cos(\beta t)\end{aligned}}}
ja
(
x
1
−
x
2
)
(
t
)
=
e
α
t
(
cos
(
β
t
)
+
i
sin
(
β
t
)
)
−
e
α
t
(
cos
(
β
t
)
−
i
sin
(
β
t
)
)
=
2
i
e
α
t
sin
(
β
t
)
{\displaystyle {\begin{aligned}(x_{1}-x_{2})(t)&=\mathrm {e} ^{\alpha t}\left(\cos(\beta t)+\mathrm {i} \sin(\beta t)\right)-\mathrm {e} ^{\alpha t}\left(\cos(\beta t)-\mathrm {i} \sin(\beta t)\right)\\&=2\mathrm {i} \mathrm {e} ^{\alpha t}\sin(\beta t)\end{aligned}}}
Merkitään
x
^
1
(
t
)
=
1
2
(
x
1
+
x
2
)
(
t
)
{\displaystyle {\hat {x}}_{1}(t)={\frac {1}{2}}(x_{1}+x_{2})(t)}
x
^
2
(
t
)
=
1
2
i
(
x
1
−
x
2
)
(
t
)
{\displaystyle {\hat {x}}_{2}(t)={\frac {1}{2\mathrm {i} }}(x_{1}-x_{2})(t)}
x
^
1
{\displaystyle {\hat {x}}_{1}}
x
^
2
{\displaystyle {\hat {x}}_{2}}
x
:
R
→
R
{\displaystyle x:\mathbb {R} \to \mathbb {R} }
x
(
t
)
=
c
1
x
^
1
(
t
)
+
c
2
x
^
2
(
t
)
=
c
1
e
α
t
cos
(
β
t
)
+
c
2
e
α
t
sin
(
β
t
)
{\displaystyle {\begin{aligned}x(t)&=c_{1}{\hat {x}}_{1}(t)+c_{2}{\hat {x}}_{2}(t)\\&=c_{1}\mathrm {e} ^{\alpha t}\cos(\beta t)+c_{2}\mathrm {e} ^{\alpha t}\sin(\beta t)\end{aligned}}}
Vakiokertoimisen differentiaaliyhtälön
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
0
{\displaystyle x''(t)+bx'(t)+cx(t)=0}
x
:
R
→
R
{\displaystyle x:\mathbb {R} \to \mathbb {R} }
Jos
b
2
−
4
c
>
0
{\displaystyle b^{2}-4c>0}
x
(
t
)
=
c
1
e
λ
1
t
+
c
2
e
λ
2
t
{\displaystyle x(t)=c_{1}\mathrm {e} ^{\lambda _{1}t}+c_{2}\mathrm {e} ^{\lambda _{2}t}}
λ
1
=
−
b
+
b
2
−
4
c
2
{\displaystyle \lambda _{1}={\frac {-b+{\sqrt {b^{2}-4c}}}{2}}}
λ
2
=
−
b
−
b
2
−
4
c
2
{\displaystyle \lambda _{2}={\frac {-b-{\sqrt {b^{2}-4c}}}{2}}}
Jos
b
2
−
4
c
=
0
{\displaystyle b^{2}-4c=0}
x
(
t
)
=
c
1
e
λ
t
+
c
2
t
e
λ
t
{\displaystyle x(t)=c_{1}\mathrm {e} ^{\lambda t}+c_{2}t\mathrm {e} ^{\lambda t}}
λ
=
−
b
2
{\displaystyle \lambda =-{\frac {b}{2}}}
Jos
b
2
−
4
c
<
0
{\displaystyle b^{2}-4c<0}
x
(
t
)
=
c
1
e
α
t
cos
(
β
t
)
+
c
2
e
α
t
sin
(
β
t
)
{\displaystyle x(t)=c_{1}\mathrm {e} ^{\alpha t}\cos(\beta t)+c_{2}\mathrm {e} ^{\alpha t}\sin(\beta t)}
α
=
−
b
2
{\displaystyle \alpha =-{\frac {b}{2}}}
β
=
4
c
−
b
2
2
{\displaystyle \beta ={\frac {\sqrt {4c-b^{2}}}{2}}}
sekä
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
Ratkaistaan differentiaaliyhtälö
x
″
+
x
′
+
x
=
0
{\displaystyle x''+x'+x=0}
λ
2
+
λ
+
1
=
0
{\displaystyle \lambda ^{2}+\lambda +1=0}
λ
=
−
1
±
1
−
4
2
=
−
1
2
±
i
3
2
{\displaystyle \lambda ={\frac {-1\pm {\sqrt {1-4}}}{2}}=-{\frac {1}{2}}\pm \mathrm {i} {\frac {\sqrt {3}}{2}}}
DY:n ratkaisu on siis
x
:
R
→
R
{\displaystyle x:\mathbb {R} \to \mathbb {R} }
x
(
t
)
=
c
1
e
−
t
/
2
cos
(
3
2
t
)
+
c
2
e
−
t
/
2
sin
(
3
2
t
)
{\displaystyle x(t)=c_{1}\mathrm {e} ^{-t/2}\cos \left({\frac {\sqrt {3}}{2}}t\right)+c_{2}\mathrm {e} ^{-t/2}\sin \left({\frac {\sqrt {3}}{2}}t\right)}
Yleinen lineaarinen differentiaaliyhtälö
muokkaa
Tässä kappaleessa osoitetaan, että yleinen lineaarinen differentiaaliyhtälö ratkeaa selvittämällä ensin vastaavan homogeenisen DY:n ratkaisukanta. Tämän jälkeen ratkaistaan DY:n epähomogeenisen osan osaratkaisut. DY:n ratkaisu on näiden kahden vaiheen summa.
Tarkastellaan lineaarista differentiaaliyhtälöä
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
(
t
)
(
♠
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q(t)\qquad (\spadesuit )}
missä
p
,
q
,
r
:
]
a
,
b
[
→
R
{\displaystyle p,q,r:\,]a,b[\,\to \mathbb {R} }
x
(
t
)
{\displaystyle x(t)}
x
^
(
t
)
{\displaystyle {\hat {x}}(t)}
♠
{\displaystyle \spadesuit }
x
1
:
]
a
,
b
[
→
R
,
x
1
(
t
)
=
x
(
t
)
−
x
^
(
t
)
{\displaystyle x_{1}:\,]a,b[\,\to \mathbb {R} ,x_{1}(t)=x(t)-{\hat {x}}(t)}
vastaavan homogeenisen DY:n
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0}
ratkaisu:
x
1
″
(
t
)
+
p
(
t
)
x
1
′
(
t
)
+
r
(
t
)
x
1
(
t
)
=
x
″
(
t
)
−
x
^
″
(
t
)
+
p
(
t
)
(
x
′
(
t
)
−
x
^
′
(
t
)
)
+
r
(
t
)
(
x
(
t
)
−
x
^
(
t
)
)
=
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
−
(
x
^
″
(
t
)
+
p
(
t
)
x
^
′
(
t
)
+
r
(
t
)
x
^
(
t
)
)
=
q
(
t
)
−
q
(
t
)
=
0
{\displaystyle {\begin{aligned}x_{1}''(t)+p(t)x_{1}'(t)+r(t)x_{1}(t)&=x''(t)-{\hat {x}}''(t)+p(t)\left(x'(t)-{\hat {x}}'(t)\right)+r(t)\left(x(t)-{\hat {x}}(t)\right)\\&=x''(t)+p(t)x'(t)+r(t)x(t)-\left({\hat {x}}''(t)+p(t){\hat {x}}'(t)+r(t){\hat {x}}(t)\right)\\&=q(t)-q(t)\\&=0\end{aligned}}}
Oletetaan, että funktiot
x
1
,
x
2
:
]
a
,
b
[
→
R
{\displaystyle x_{1},x_{2}:\,]a,b[\,\to \mathbb {R} }
x
3
:
]
a
,
b
[
→
R
{\displaystyle x_{3}:\,]a,b[\,\to \mathbb {R} }
♠
{\displaystyle \spadesuit }
♠
{\displaystyle \spadesuit }
kaikki ratkaisut ovat muotoa
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
+
x
3
(
t
)
{\displaystyle x(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)+x_{3}(t)}
Lauseen 6 todistus
Olkoon
x
:
]
a
,
b
[
→
R
{\displaystyle x:\,]a,b[\,\to \mathbb {R} }
♠
{\displaystyle \spadesuit }
w
:
]
a
,
b
[
→
R
,
w
(
t
)
=
x
(
t
)
−
x
3
(
t
)
{\displaystyle w:\,]a,b[\,\to \mathbb {R} ,w(t)=x(t)-x_{3}(t)}
w
{\displaystyle w}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
ratkaisukannan , joten on olemassa vakiot
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
w
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
{\displaystyle w(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
x
(
t
)
−
x
3
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
{\displaystyle x(t)-x_{3}(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)}
kaikilla
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Q.E.D.
HUOM! Jos
x
1
{\displaystyle x_{1}}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
1
(
t
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q_{1}(t)}
x
2
{\displaystyle x_{2}}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
2
(
t
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q_{2}(t)}
x
1
+
x
2
{\displaystyle x_{1}+x_{2}}
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
1
(
t
)
+
q
2
(
t
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q_{1}(t)+q_{2}(t)}
Ratkaistaan differentiaaliyhtälö
x
″
+
4
x
=
1
+
t
+
sin
t
{\displaystyle x''+4x=1+t+\sin t}
x
″
+
4
x
=
0
{\displaystyle x''+4x=0}
λ
2
+
4
=
0
{\displaystyle \lambda ^{2}+4=0}
λ
=
±
2
i
{\displaystyle \lambda =\pm 2\mathrm {i} }
Lauseen 5 mukaan homogeenisen DY:n ratkaisu on
c
1
cos
(
2
t
)
+
c
2
sin
(
2
t
)
{\displaystyle c_{1}\cos(2t)+c_{2}\sin(2t)}
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
Etsitään epähomogeenisen osan osaratkaisut:
x
″
(
t
)
+
4
x
(
t
)
=
1
⇒
x
(
t
)
≡
1
4
x
″
(
t
)
+
4
x
(
t
)
=
t
⇒
x
(
t
)
=
1
4
t
x
″
(
t
)
+
4
x
(
t
)
=
sin
t
⇒
x
(
t
)
=
1
3
sin
t
{\displaystyle {\begin{aligned}x''(t)+4x(t)=1\quad &\Rightarrow \quad x(t)\equiv {\frac {1}{4}}\\x''(t)+4x(t)=t\quad &\Rightarrow \quad x(t)={\frac {1}{4}}t\\x''(t)+4x(t)=\sin t\quad &\Rightarrow \quad x(t)={\frac {1}{3}}\sin t\end{aligned}}}
DY:n ratkaisu on siis
x
:
R
→
R
,
x
(
t
)
=
c
1
cos
(
2
t
)
+
c
2
sin
(
2
t
)
+
1
3
sin
t
+
1
4
t
+
1
4
{\displaystyle x:\mathbb {R} \to \mathbb {R} ,x(t)=c_{1}\cos(2t)+c_{2}\sin(2t)+{\frac {1}{3}}\sin t+{\frac {1}{4}}t+{\frac {1}{4}}}
Vakiokertoiminen ei-homogeeninen differentiaaliyhtälö
muokkaa
Mikäli vakiokertoimisessa lineaarisessa differentiaaliyhtälössä
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
q
(
t
)
{\textstyle x''(t)+bx'(t)+cx(t)=q(t)}
q
{\displaystyle q}
sivistyneeksi arvaukseksi .
Jos q on polynomi
muokkaa
Olkoon
q
{\displaystyle q}
n . asteen polynomi . Ts. olkoon
n
∈
N
{\displaystyle n\in \mathbb {N} }
a
0
,
a
1
,
a
2
,
…
,
a
n
∈
R
{\displaystyle a_{0},a_{1},a_{2},\dots ,a_{n}\in \mathbb {R} }
q
:
R
→
R
{\displaystyle q:\mathbb {R} \to \mathbb {R} }
q
(
t
)
=
a
n
t
n
+
a
n
−
1
t
n
−
1
+
⋯
+
a
2
t
2
+
a
1
t
+
a
0
=
∑
k
=
0
n
a
k
t
k
{\displaystyle q(t)=a_{n}t^{n}+a_{n-1}t^{n-1}+\dots +a_{2}t^{2}+a_{1}t+a_{0}=\sum _{k=0}^{n}a_{k}t^{k}}
Arvataan, että epähomogeenisen osan osaratkaisu
x
3
{\displaystyle x_{3}}
n . asteen polynomi. Olkoon
A
0
,
A
1
,
A
2
,
…
,
A
n
∈
R
{\displaystyle A_{0},A_{1},A_{2},\dots ,A_{n}\in \mathbb {R} }
x
3
(
t
)
=
A
n
t
n
+
A
n
−
1
t
n
−
1
+
⋯
+
A
2
t
2
+
A
1
t
+
A
0
=
∑
k
=
0
n
A
k
t
k
{\displaystyle x_{3}(t)=A_{n}t^{n}+A_{n-1}t^{n-1}+\dots +A_{2}t^{2}+A_{1}t+A_{0}=\sum _{k=0}^{n}A_{k}t^{k}}
Tällöin
x
3
′
(
t
)
=
n
A
n
t
n
−
1
+
(
n
−
1
)
A
n
−
1
t
n
−
2
+
⋯
+
2
A
2
t
+
A
1
=
∑
k
=
1
n
k
A
k
t
k
−
1
x
3
″
(
t
)
=
n
(
n
−
1
)
A
n
t
n
−
2
+
(
n
−
1
)
(
n
−
2
)
A
n
−
1
t
n
−
3
+
⋯
+
2
A
2
=
∑
k
=
2
n
k
(
k
−
1
)
A
k
t
k
−
2
{\displaystyle {\begin{aligned}x_{3}'(t)&=nA_{n}t^{n-1}+(n-1)A_{n-1}t^{n-2}+\dots +2A_{2}t+A_{1}=\sum _{k=1}^{n}kA_{k}t^{k-1}\\x_{3}''(t)&=n(n-1)A_{n}t^{n-2}+(n-1)(n-2)A_{n-1}t^{n-3}+\dots +2A_{2}=\sum _{k=2}^{n}k(k-1)A_{k}t^{k-2}\end{aligned}}}
Sijoitetaan derivoidut lausekkeet differentiaaliyhtälöön:
x
3
″
(
t
)
+
b
x
3
′
(
t
)
+
c
x
3
(
t
)
=
q
(
t
)
∑
k
=
2
n
k
(
k
−
1
)
A
k
t
k
−
2
+
b
∑
k
=
1
n
k
A
k
t
k
−
1
+
c
∑
k
=
0
n
A
k
t
k
=
∑
k
=
0
n
a
k
t
k
{\displaystyle {\begin{aligned}x_{3}''(t)+bx_{3}'(t)+cx_{3}(t)&=q(t)\\\sum _{k=2}^{n}k(k-1)A_{k}t^{k-2}+b\sum _{k=1}^{n}kA_{k}t^{k-1}+c\sum _{k=0}^{n}A_{k}t^{k}&=\sum _{k=0}^{n}a_{k}t^{k}\end{aligned}}}
Yhtälön kummallakin puolella jokaisen asteen monomien kertoimien on oltava yhtä suuret. Saadaan yhtälöryhmä:
{
c
A
n
t
n
=
a
n
t
n
(
b
n
A
n
+
c
A
n
−
1
)
t
n
−
1
=
a
n
−
1
t
n
−
1
(
n
(
n
−
1
)
A
n
+
b
(
n
−
1
)
A
n
−
1
+
c
A
n
−
2
)
t
n
−
2
=
a
n
−
2
t
n
−
2
(
(
n
−
1
)
(
n
−
2
)
A
n
−
1
+
b
(
n
−
2
)
A
n
−
2
+
c
A
n
−
3
)
t
n
−
3
=
a
n
−
3
t
n
−
3
⋮
(
4
⋅
3
A
4
+
3
b
A
3
+
c
A
2
)
t
2
=
a
2
t
2
(
3
⋅
2
A
3
+
2
b
A
2
+
c
A
1
)
t
=
a
1
t
2
A
2
+
b
A
1
+
c
A
0
=
a
0
{\displaystyle {\begin{cases}cA_{n}t^{n}&=a_{n}t^{n}\\\left(bnA_{n}+cA_{n-1}\right)t^{n-1}&=a_{n-1}t^{n-1}\\\left(n(n-1)A_{n}+b(n-1)A_{n-1}+cA_{n-2}\right)t^{n-2}&=a_{n-2}t^{n-2}\\\left((n-1)(n-2)A_{n-1}+b(n-2)A_{n-2}+cA_{n-3}\right)t^{n-3}&=a_{n-3}t^{n-3}\\&\vdots \\\left(4\cdot 3A_{4}+3bA_{3}+cA_{2}\right)t^{2}&=a_{2}t^{2}\\\left(3\cdot 2A_{3}+2bA_{2}+cA_{1}\right)t&=a_{1}t\\2A_{2}+bA_{1}+cA_{0}&=a_{0}\end{cases}}}
Toisin sanoen
{
c
A
n
=
a
n
b
n
A
n
+
c
A
n
−
1
=
a
n
−
1
n
(
n
−
1
)
A
n
+
b
(
n
−
1
)
A
n
−
1
+
c
A
n
−
2
=
a
n
−
2
(
n
−
1
)
(
n
−
2
)
A
n
−
1
+
b
(
n
−
2
)
A
n
−
2
+
c
A
n
−
3
=
a
n
−
3
⋮
4
⋅
3
A
4
+
3
b
A
3
+
c
A
2
=
a
2
3
⋅
2
A
3
+
2
b
A
2
+
c
A
1
=
a
1
2
A
2
+
b
A
1
+
c
A
0
=
a
0
{\displaystyle {\begin{cases}cA_{n}&=a_{n}\\bnA_{n}+cA_{n-1}&=a_{n-1}\\n(n-1)A_{n}+b(n-1)A_{n-1}+cA_{n-2}&=a_{n-2}\\(n-1)(n-2)A_{n-1}+b(n-2)A_{n-2}+cA_{n-3}&=a_{n-3}\\&\vdots \\4\cdot 3A_{4}+3bA_{3}+cA_{2}&=a_{2}\\3\cdot 2A_{3}+2bA_{2}+cA_{1}&=a_{1}\\2A_{2}+bA_{1}+cA_{0}&=a_{0}\end{cases}}}
Jäljelle jää ratkaistavaksi yhtälöryhmästä kertoimet
A
0
,
A
1
,
A
2
,
…
,
A
n
{\displaystyle A_{0},A_{1},A_{2},\dots ,A_{n}}
Ratkaistaan differentiaaliyhtälön
x
″
−
3
x
′
−
4
x
=
4
t
2
+
2
{\displaystyle x''-3x'-4x=4t^{2}+2}
q
(
t
)
=
4
t
2
+
2
{\displaystyle q(t)=4t^{2}+2}
x
3
(
t
)
=
A
2
t
2
+
A
1
t
+
A
0
{\displaystyle x_{3}(t)=A_{2}t^{2}+A_{1}t+A_{0}}
x
3
′
(
t
)
=
2
A
2
t
+
A
1
x
3
″
(
t
)
=
2
A
2
{\displaystyle {\begin{aligned}x_{3}'(t)&=2A_{2}t+A_{1}\\x_{3}''(t)&=2A_{2}\end{aligned}}}
Sijoitetaan nämä DY:öön:
2
A
2
−
3
(
2
A
2
t
+
A
1
)
−
4
(
A
2
t
2
+
A
1
t
+
A
0
)
=
4
t
2
+
2
−
4
A
2
t
2
−
(
6
A
2
+
4
A
1
)
t
+
2
A
2
−
3
A
1
−
4
A
0
=
4
t
2
+
2
{\displaystyle {\begin{aligned}2A_{2}-3\left(2A_{2}t+A_{1}\right)-4\left(A_{2}t^{2}+A_{1}t+A_{0}\right)&=4t^{2}+2\\-4A_{2}t^{2}-\left(6A_{2}+4A_{1}\right)t+2A_{2}-3A_{1}-4A_{0}&=4t^{2}+2\end{aligned}}}
Saadaan yhtälöryhmä
{
−
4
A
2
=
4
−
6
A
2
−
4
A
1
=
0
2
A
2
−
3
A
1
−
4
A
0
=
2
{\displaystyle {\begin{cases}-4A_{2}&=4\\-6A_{2}-4A_{1}&=0\\2A_{2}-3A_{1}-4A_{0}&=2\end{cases}}}
Yhtälöryhmän ratkaisu on
{
A
2
=
−
1
A
1
=
3
2
A
0
=
−
17
8
{\displaystyle {\begin{cases}A_{2}=-1\\A_{1}={\frac {3}{2}}\\A_{0}=-{\frac {17}{8}}\end{cases}}}
Epähomogeenisen osan ratkaisu on siis
x
3
(
t
)
=
−
t
2
+
3
2
t
−
17
8
{\displaystyle x_{3}(t)=-t^{2}+{\frac {3}{2}}t-{\frac {17}{8}}}
Jos q on eksponenttifunktio
muokkaa
Olkoon
q
{\displaystyle q}
A
,
r
∈
R
{\displaystyle A,r\in \mathbb {R} }
q
:
R
→
R
,
q
(
t
)
=
A
e
r
t
{\displaystyle q:\mathbb {R} \to \mathbb {R} ,~q(t)=A\mathrm {e} ^{rt}}
x
3
{\displaystyle x_{3}}
karakteristisen yhtälön (
♡
{\displaystyle \heartsuit }
r
{\displaystyle r}
Jos
r
{\displaystyle r}
ei ole karakteristisen yhtälön juuri , niin epähomogeenisen osan ratkaisu on
x
3
(
t
)
=
K
e
r
t
{\displaystyle x_{3}(t)=K\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Jos
r
{\displaystyle r}
on karakteristisen yhtälön toinen juuri (
b
2
−
4
c
≠
0
{\displaystyle b^{2}-4c\neq 0}
) , niin epähomogeenisen osan ratkaisu on
x
3
(
t
)
=
K
t
e
r
t
{\displaystyle x_{3}(t)=Kt\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Jos
r
{\displaystyle r}
karakteristisen yhtälön kaksinkertainen juuri (
b
2
−
4
c
=
0
{\displaystyle b^{2}-4c=0}
, niin epähomogeenisen osan ratkaisu on
x
3
(
t
)
=
K
t
2
e
r
t
{\displaystyle x_{3}(t)=Kt^{2}\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Ratkaistaan seuraavien differentiaaliyhtälöiden epähomogeeniset osat: a)
x
″
+
2
x
′
+
3
x
=
2
e
−
2
t
{\displaystyle x''+2x'+3x={\sqrt {2}}\mathrm {e} ^{-2t}}
b)
x
″
−
4
x
=
2
e
2
t
{\displaystyle x''-4x={\sqrt {2}}\mathrm {e} ^{2t}}
c)
x
″
+
2
x
′
+
x
=
2
e
−
t
{\displaystyle x''+2x'+x={\sqrt {2}}\mathrm {e} ^{-t}}
a) DY:n karakteristinen yhtälö on
λ
2
+
2
λ
+
3
=
0
{\displaystyle \lambda ^{2}+2\lambda +3=0}
λ
1
=
−
1
+
i
2
{\displaystyle \lambda _{1}=-1+\mathrm {i} {\sqrt {2}}}
λ
2
=
−
1
−
i
2
{\displaystyle \lambda _{2}=-1-\mathrm {i} {\sqrt {2}}}
r
=
−
2
{\displaystyle r=-2}
x
3
(
t
)
=
K
e
r
t
=
K
e
−
2
t
{\displaystyle x_{3}(t)=K\mathrm {e} ^{rt}=K\mathrm {e} ^{-2t}}
K
{\displaystyle K}
x
3
″
(
t
)
+
2
x
3
′
(
t
)
+
3
x
3
(
t
)
=
2
e
−
2
t
4
K
e
−
2
t
+
2
⋅
(
−
2
K
e
−
2
t
)
+
3
K
e
−
2
t
=
2
e
−
2
t
3
K
e
−
2
t
=
2
e
−
2
t
|
|
:
3
e
−
2
t
(
≠
0
∀
t
∈
R
)
K
=
2
3
{\displaystyle {\begin{aligned}x_{3}''(t)+2x_{3}'(t)+3x_{3}(t)&={\sqrt {2}}\mathrm {e} ^{-2t}\\4K\mathrm {e} ^{-2t}+2\cdot \left(-2K\mathrm {e} ^{-2t}\right)+3K\mathrm {e} ^{-2t}&={\sqrt {2}}\mathrm {e} ^{-2t}\\3K\mathrm {e} ^{-2t}&={\sqrt {2}}\mathrm {e} ^{-2t}\quad ||~:3\mathrm {e} ^{-2t}~(\neq 0~\forall ~t\in \mathbb {R} )\\K&={\frac {\sqrt {2}}{3}}\end{aligned}}}
Epähomogeenisen osan ratkaisu on siis
x
3
(
t
)
=
2
3
e
−
2
t
{\displaystyle x_{3}(t)={\frac {\sqrt {2}}{3}}\mathrm {e} ^{-2t}}
b) DY:n karakteristinen yhtälö on
λ
2
−
4
=
0
{\displaystyle \lambda ^{2}-4=0}
λ
1
=
2
{\displaystyle \lambda _{1}=2}
λ
2
=
−
2
{\displaystyle \lambda _{2}=-2}
r
=
2
=
λ
1
{\displaystyle r=2=\lambda _{1}}
x
3
(
t
)
=
K
t
e
r
t
=
K
t
e
2
t
{\displaystyle x_{3}(t)=Kt\mathrm {e} ^{rt}=Kt\mathrm {e} ^{2t}}
K
{\displaystyle K}
x
3
″
(
t
)
−
4
x
3
(
t
)
=
2
e
2
t
K
(
4
e
2
t
+
4
t
e
2
t
)
−
4
K
t
e
2
t
=
2
e
2
t
4
K
e
2
t
=
2
e
2
t
|
|
:
4
e
2
t
K
=
2
4
=
1
2
2
{\displaystyle {\begin{aligned}x_{3}''(t)-4x_{3}(t)&={\sqrt {2}}\mathrm {e} ^{2t}\\K(4\mathrm {e} ^{2t}+4t\mathrm {e} ^{2t})-4Kt\mathrm {e} ^{2t}&={\sqrt {2}}\mathrm {e} ^{2t}\\4K\mathrm {e} ^{2t}&={\sqrt {2}}\mathrm {e} ^{2t}\quad ||~:\mathrm {4} e^{2t}\\K&={\frac {\sqrt {2}}{4}}={\frac {1}{2{\sqrt {2}}}}\end{aligned}}}
Epähomogeenisen osan ratkaisu on siis
x
3
(
t
)
=
t
e
−
2
t
2
2
{\displaystyle x_{3}(t)={\frac {t\mathrm {e} ^{-2t}}{2{\sqrt {2}}}}}
c) DY:n karakteristinen yhtälö on
λ
2
+
2
λ
+
1
=
0
{\displaystyle \lambda ^{2}+2\lambda +1=0}
λ
=
−
1
{\displaystyle \lambda =-1}
r
=
−
1
=
λ
{\displaystyle r=-1=\lambda }
x
3
(
t
)
=
K
t
2
e
r
t
=
K
t
2
e
−
t
{\displaystyle x_{3}(t)=Kt^{2}\mathrm {e} ^{rt}=Kt^{2}\mathrm {e} ^{-t}}
K
{\displaystyle K}
x
3
″
(
t
)
+
2
x
3
′
(
t
)
+
x
3
(
t
)
=
2
e
−
t
K
e
−
t
(
t
2
−
4
t
+
2
)
+
2
K
e
−
t
(
−
t
2
+
2
t
)
+
K
t
2
e
−
t
=
2
e
−
t
2
K
e
−
t
=
2
e
−
t
|
|
:
2
e
−
t
K
=
2
2
=
1
2
{\displaystyle {\begin{aligned}x_{3}''(t)+2x_{3}'(t)+x_{3}(t)&={\sqrt {2}}\mathrm {e} ^{-t}\\K\mathrm {e} ^{-t}(t^{2}-4t+2)+2K\mathrm {e} ^{-t}(-t^{2}+2t)+Kt^{2}\mathrm {e} ^{-t}&={\sqrt {2}}\mathrm {e} ^{-t}\\2K\mathrm {e} ^{-t}&={\sqrt {2}}\mathrm {e} ^{-t}\quad ||~:2\mathrm {e} ^{-t}\\K&={\frac {\sqrt {2}}{2}}={\frac {1}{\sqrt {2}}}\end{aligned}}}
Epähomogeenisen osan ratkaisu on siis
x
3
(
t
)
=
t
2
e
−
2
t
2
{\displaystyle x_{3}(t)={\frac {t^{2}\mathrm {e} ^{-2t}}{\sqrt {2}}}}
Jos q on sini- tai kosinifunktio
muokkaa
Sekä sini- että kosinifunktio käsitellään samalla tavalla. Koska
sin
θ
=
cos
(
π
2
−
θ
)
{\displaystyle \sin \theta =\cos \left({\frac {\pi }{2}}-\theta \right)}
θ
∈
R
{\displaystyle \theta \in \mathbb {R} }
q
{\displaystyle q}
A
,
ω
∈
R
{\displaystyle A,\omega \in \mathbb {R} }
q
:
R
→
R
,
q
(
t
)
=
A
cos
(
ω
t
)
{\displaystyle q:\mathbb {R} \to \mathbb {R} ,~q(t)=A\cos(\omega t)}
x
3
{\displaystyle x_{3}}
♡
{\displaystyle \heartsuit }
ω
{\displaystyle \omega }
Jos
ω
i
{\displaystyle \omega \mathrm {i} }
ei ole karakteristisen yhtälön juuri , niin epähomogeenisen osan ratkaisu on
x
3
(
t
)
=
K
sin
(
ω
t
)
+
L
cos
(
ω
t
)
{\displaystyle x_{3}(t)=K\sin(\omega t)+L\cos(\omega t)}
K
,
L
∈
R
{\displaystyle K,L\in \mathbb {R} }
Jos
ω
i
{\displaystyle \omega \mathrm {i} }
on karakteristisen yhtälön juuri , niin epähomogeenisen osan ratkaisu on
x
3
(
t
)
=
K
t
sin
(
ω
t
)
{\displaystyle x_{3}(t)=Kt\sin(\omega t)}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Ratkaistaan seuraavien differentiaaliyhtälöiden epähomogeeniset osat: a)
x
″
+
2
x
′
+
2
x
=
5
sin
t
{\displaystyle x''+2x'+2x=5\sin t}
b)
x
″
+
4
x
=
cos
(
2
t
)
{\displaystyle x''+4x=\cos(2t)}
a) Tehdään aluksi muuttujanvaihto
s
=
π
2
−
t
{\displaystyle s={\frac {\pi }{2}}-t}
x
″
+
2
x
′
+
2
x
=
5
cos
s
{\displaystyle x''+2x'+2x=5\cos s}
λ
2
+
2
λ
+
2
=
0
{\displaystyle \lambda ^{2}+2\lambda +2=0}
λ
1
=
−
1
+
i
{\displaystyle \lambda _{1}=-1+\mathrm {i} }
λ
2
=
−
1
−
i
{\displaystyle \lambda _{2}=-1-\mathrm {i} }
ω
=
1
⇒
ω
i
=
i
{\displaystyle \omega =1\Rightarrow \omega \mathrm {i} =\mathrm {i} }
x
3
(
s
)
=
K
sin
(
ω
s
)
+
L
cos
(
ω
s
)
=
K
sin
s
+
L
cos
s
=
K
sin
(
π
2
−
t
)
+
L
cos
(
π
2
−
t
)
=
K
cos
t
+
L
sin
t
{\displaystyle {\begin{aligned}x_{3}(s)&=K\sin(\omega s)+L\cos(\omega s)=K\sin s+L\cos s\\&=K\sin \left({\frac {\pi }{2}}-t\right)+L\cos \left({\frac {\pi }{2}}-t\right)\\&=K\cos t+L\sin t\end{aligned}}}
Sijoitetaan tämä DY:öön ja ratkaistaan vakiot
K
{\displaystyle K}
L
{\displaystyle L}
x
3
″
(
t
)
+
2
x
3
′
(
t
)
+
2
x
3
(
t
)
=
5
sin
t
−
K
cos
t
−
L
sin
t
−
2
K
sin
t
+
2
L
cos
t
+
2
K
cos
t
+
2
L
sin
t
=
5
sin
t
(
−
2
K
+
L
)
sin
t
+
(
K
+
2
L
)
cos
t
=
5
sin
t
{\displaystyle {\begin{aligned}x_{3}''(t)+2x_{3}'(t)+2x_{3}(t)&=5\sin t\\-K\cos t-L\sin t-2K\sin t+2L\cos t+2K\cos t+2L\sin t&=5\sin t\\(-2K+L)\sin t+(K+2L)\cos t&=5\sin t\end{aligned}}}
Saadaan yhtälöpari:
{
−
2
K
+
L
=
5
K
+
2
L
=
0
⇔
{
K
=
−
2
L
=
1
{\displaystyle {\begin{cases}-2K+L=5\\K+2L=0\end{cases}}\Leftrightarrow {\begin{cases}K=-2\\L=1\end{cases}}}
Epähomogeenisen osan ratkaisu on siis
x
3
(
t
)
=
K
cos
t
+
L
sin
t
=
−
2
cos
t
+
sin
t
{\displaystyle x_{3}(t)=K\cos t+L\sin t=-2\cos t+\sin t}
b) DY:n karakteristinen yhtälö on
λ
2
+
4
=
0
{\displaystyle \lambda ^{2}+4=0}
λ
1
=
2
i
{\displaystyle \lambda _{1}=2\mathrm {i} }
λ
2
=
−
2
i
{\displaystyle \lambda _{2}=-2\mathrm {i} }
ω
=
2
⇒
ω
i
=
2
i
=
λ
1
{\displaystyle \omega =2\Rightarrow \omega \mathrm {i} =2\mathrm {i} =\lambda _{1}}
x
3
(
t
)
=
K
t
sin
(
ω
t
)
=
K
t
sin
(
2
t
)
{\displaystyle x_{3}(t)=Kt\sin(\omega t)=Kt\sin(2t)}
K
{\displaystyle K}
x
3
″
(
t
)
+
4
x
3
(
t
)
=
cos
(
2
t
)
K
(
4
cos
(
2
t
)
−
4
t
sin
(
2
t
)
)
+
4
K
t
sin
(
2
t
)
=
cos
(
2
t
)
4
K
cos
(
2
t
)
=
cos
(
2
t
)
⇔
K
=
1
4
{\displaystyle {\begin{aligned}x_{3}''(t)+4x_{3}(t)&=\cos(2t)\\K\left(4\cos(2t)-4t\sin(2t)\right)+4Kt\sin(2t)&=\cos(2t)\\4K\cos(2t)&=\cos(2t)\\\Leftrightarrow K&={\frac {1}{4}}\end{aligned}}}
x
3
(
t
)
=
1
4
t
sin
(
2
t
)
{\displaystyle x_{3}(t)={\frac {1}{4}}t\sin(2t)}
Vakiokertoimisen differentiaaliyhtälön
x
″
(
t
)
+
b
x
′
(
t
)
+
c
x
(
t
)
=
q
(
t
)
{\displaystyle x''(t)+bx'(t)+cx(t)=q(t)}
x
3
:
R
→
R
{\displaystyle x_{3}:\mathbb {R} \to \mathbb {R} }
Jos
q
{\displaystyle q}
n . asteen polynomi, niin
x
3
{\displaystyle x_{3}}
Jos
c
≠
0
{\displaystyle c\neq 0}
x
3
(
t
)
=
∑
k
=
0
n
A
k
t
k
{\displaystyle x_{3}(t)=\sum _{k=0}^{n}A_{k}t^{k}}
x
3
{\displaystyle x_{3}}
n. asteen polynomi.
Jos
c
=
0
{\displaystyle c=0}
b
≠
0
{\displaystyle b\neq 0}
x
3
(
t
)
=
t
∑
k
=
0
n
A
k
t
k
{\displaystyle x_{3}(t)=t\sum _{k=0}^{n}A_{k}t^{k}}
x
3
{\displaystyle x_{3}}
n + 1. asteen polynomi.
Jos
q
(
t
)
=
A
e
r
t
{\displaystyle q(t)=A\mathrm {e} ^{rt}}
Jos
r
{\displaystyle r}
x
3
(
t
)
=
K
e
r
t
{\displaystyle x_{3}(t)=K\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Jos
r
{\displaystyle r}
x
3
(
t
)
=
K
t
e
r
t
{\displaystyle x_{3}(t)=Kt\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Jos
r
{\displaystyle r}
x
3
(
t
)
=
K
t
2
e
r
t
{\displaystyle x_{3}(t)=Kt^{2}\mathrm {e} ^{rt}}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Jos
q
(
t
)
=
A
cos
(
ω
t
)
{\displaystyle q(t)=A\cos(\omega t)}
Jos
ω
i
{\displaystyle \omega \mathrm {i} }
x
3
(
t
)
=
K
sin
(
ω
t
)
+
L
cos
(
ω
t
)
{\displaystyle x_{3}(t)=K\sin(\omega t)+L\cos(\omega t)}
K
,
L
∈
R
{\displaystyle K,L\in \mathbb {R} }
Jos
ω
i
{\displaystyle \omega \mathrm {i} }
x
3
(
t
)
=
K
t
sin
(
ω
t
)
{\displaystyle x_{3}(t)=Kt\sin(\omega t)}
K
∈
R
{\displaystyle K\in \mathbb {R} }
Vakioiden variointi
muokkaa
Tarkastellaan jälleen yleistä lineaarista differentiaaliyhtälöä
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
q
(
t
)
(
♠
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=q(t)\quad (\spadesuit )}
missä
p
,
q
,
r
:
]
a
,
b
[
→
R
{\displaystyle p,q,r:\,]a,b[\,\to \mathbb {R} }
x
1
,
x
2
:
]
a
,
b
[
→
R
{\displaystyle x_{1},x_{2}:\,]a,b[\,\to \mathbb {R} }
x
″
(
t
)
+
p
(
t
)
x
′
(
t
)
+
r
(
t
)
x
(
t
)
=
0
(
♢
)
{\displaystyle x''(t)+p(t)x'(t)+r(t)x(t)=0\quad (\diamondsuit )}
ratkaisukannan. Kuten edellä todettiin , DY:n (
♢
{\displaystyle \diamondsuit }
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
{\displaystyle x(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)}
♠
{\displaystyle \spadesuit }
x
(
t
)
=
c
1
x
1
(
t
)
+
c
2
x
2
(
t
)
+
x
3
(
t
)
{\displaystyle x(t)=c_{1}x_{1}(t)+c_{2}x_{2}(t)+x_{3}(t)}
x
3
:
]
a
,
b
[
→
R
{\displaystyle x_{3}:\,]a,b[\,\to \mathbb {R} }
♠
{\displaystyle \spadesuit }
x
3
{\displaystyle x_{3}}
vakioiden varioinnilla seuraavasti:
Olkoon
u
1
,
u
2
:
]
a
,
b
[
→
R
{\displaystyle u_{1},u_{2}:\,]a,b[\,\to \mathbb {R} }
x
(
t
)
=
u
1
(
t
)
x
1
(
t
)
+
u
2
(
t
)
x
2
(
t
)
{\displaystyle x(t)=u_{1}(t)x_{1}(t)+u_{2}(t)x_{2}(t)}
Tarkoituksena on ratkaista funktiot
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
♠
{\displaystyle \spadesuit }
x
{\displaystyle x}
x
′
(
t
)
=
u
1
′
(
t
)
x
1
(
t
)
+
u
1
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
(
t
)
+
u
2
(
t
)
x
2
′
(
t
)
=
(
u
1
′
x
1
(
t
)
+
u
2
′
(
t
)
x
2
(
t
)
)
+
(
u
1
(
t
)
x
1
′
(
t
)
+
u
2
(
t
)
x
2
′
(
t
)
)
{\displaystyle {\begin{aligned}x'(t)&=u_{1}'(t)x_{1}(t)+u_{1}(t)x_{1}'(t)+u_{2}'(t)x_{2}(t)+u_{2}(t)x_{2}'(t)\\&=\left(u_{1}'x_{1}(t)+u_{2}'(t)x_{2}(t)\right)+\left(u_{1}(t)x_{1}'(t)+u_{2}(t)x_{2}'(t)\right)\end{aligned}}}
Valitaan
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
1)
u
1
′
(
t
)
x
1
(
t
)
+
u
2
′
(
t
)
x
2
(
t
)
=
0
{\displaystyle u_{1}'(t)x_{1}(t)+u_{2}'(t)x_{2}(t)=0}
x
″
(
t
)
=
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
+
u
1
(
t
)
x
1
″
(
t
)
+
u
2
(
t
)
x
2
″
(
t
)
{\displaystyle x''(t)=u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)+u_{1}(t)x_{1}''(t)+u_{2}(t)x_{2}''(t)}
Sijoitetaan nämä tiedot DY:öön (
♠
{\displaystyle \spadesuit }
u
1
′
(
t
)
x
1
″
(
t
)
+
u
2
(
t
)
x
2
″
(
t
)
+
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
+
p
(
t
)
(
u
1
(
t
)
x
1
′
(
t
)
+
u
2
(
t
)
x
2
′
(
t
)
)
+
r
(
t
)
(
u
1
(
t
)
x
1
(
t
)
+
u
2
(
t
)
x
2
(
t
)
)
=
q
(
t
)
{\displaystyle u_{1}'(t)x_{1}''(t)+u_{2}(t)x_{2}''(t)+u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)+p(t)\left(u_{1}(t)x_{1}'(t)+u_{2}(t)x_{2}'(t)\right)+r(t)\left(u_{1}(t)x_{1}(t)+u_{2}(t)x_{2}(t)\right)=q(t)}
Järjestellään termejä siten, että
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
u
1
(
t
)
(
x
1
″
(
t
)
+
p
(
t
)
x
1
′
(
t
)
+
r
(
t
)
x
1
(
t
)
)
+
u
2
(
t
)
(
x
2
″
(
t
)
+
p
(
t
)
x
2
′
(
t
)
+
r
(
t
)
x
2
(
t
)
)
+
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
=
q
(
t
)
{\displaystyle u_{1}(t)\left(x_{1}''(t)+p(t)x_{1}'(t)+r(t)x_{1}(t)\right)+u_{2}(t)\left(x_{2}''(t)+p(t)x_{2}'(t)+r(t)x_{2}(t)\right)+u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)=q(t)}
Toisaalta, koska funktiot
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
♢
{\displaystyle \diamondsuit }
u
1
(
t
)
(
x
1
″
(
t
)
+
p
(
t
)
x
1
′
(
t
)
+
r
(
t
)
x
1
(
t
)
)
⏟
=
0
+
u
2
(
t
)
(
x
2
″
(
t
)
+
p
(
t
)
x
2
′
(
t
)
+
r
(
t
)
x
2
(
t
)
)
⏟
=
0
+
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
=
q
(
t
)
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
=
q
(
t
)
{\displaystyle {\begin{aligned}u_{1}(t)\underbrace {\left(x_{1}''(t)+p(t)x_{1}'(t)+r(t)x_{1}(t)\right)} _{=0}+u_{2}(t)\underbrace {\left(x_{2}''(t)+p(t)x_{2}'(t)+r(t)x_{2}(t)\right)} _{=0}+u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)&=q(t)\\u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)&=q(t)\end{aligned}}}
2) Nyt pitää olla
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
=
q
(
t
)
{\displaystyle u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)=q(t)}
♠
{\displaystyle \spadesuit }
Ehdoista 1) ja 2) saadaan yhtälöpari:
{
u
1
′
(
t
)
x
1
(
t
)
+
u
2
′
(
t
)
x
2
(
t
)
=
0
u
1
′
(
t
)
x
1
′
(
t
)
+
u
2
′
(
t
)
x
2
′
(
t
)
=
q
(
t
)
{\displaystyle {\begin{cases}u_{1}'(t)x_{1}(t)+u_{2}'(t)x_{2}(t)=0\\u_{1}'(t)x_{1}'(t)+u_{2}'(t)x_{2}'(t)=q(t)\end{cases}}}
Yhtälöparista ratkaistaan derivaatat
u
1
′
{\displaystyle u_{1}'}
u
2
′
{\displaystyle u_{2}'}
{
u
1
′
(
t
)
=
−
x
2
(
t
)
q
(
t
)
W
(
x
1
,
x
2
)
(
t
)
(
♣
1
)
u
2
′
(
t
)
=
x
1
(
t
)
q
(
t
)
W
(
x
1
,
x
2
)
(
t
)
(
♣
2
)
{\displaystyle {\begin{cases}\displaystyle {u_{1}'(t)=-{\frac {x_{2}(t)q(t)}{W(x_{1},x_{2})(t)}}}\quad (\clubsuit 1)\\\displaystyle {u_{2}'(t)={\frac {x_{1}(t)q(t)}{W(x_{1},x_{2})(t)}}}\quad (\clubsuit 2)\end{cases}}}
Funktiot
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
♣
1
{\displaystyle \clubsuit 1}
♣
2
{\displaystyle \clubsuit 2}
x
3
(
t
)
=
u
1
(
t
)
x
1
(
t
)
+
u
2
(
t
)
x
2
(
t
)
{\displaystyle x_{3}(t)=u_{1}(t)x_{1}(t)+u_{2}(t)x_{2}(t)}
HUOM! Lausekkeet (
♣
1
{\displaystyle \clubsuit 1}
♣
2
{\displaystyle \clubsuit 2}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
W
(
x
1
,
x
2
)
(
t
)
≠
0
{\displaystyle W(x_{1},x_{2})(t)\neq 0}
t
∈
]
a
,
b
[
{\displaystyle t\in \,]a,b[}
Ratkaistaan differentiaaliyhtälö
x
″
(
t
)
+
x
(
t
)
=
1
cos
t
,
0
<
t
<
π
2
{\displaystyle x''(t)+x(t)={\frac {1}{\cos t}},\qquad 0<t<{\frac {\pi }{2}}}
Vastaava homogeeninen DY on
x
″
(
t
)
+
x
(
t
)
=
0
{\displaystyle x''(t)+x(t)=0}
λ
2
+
1
=
0
{\displaystyle \lambda ^{2}+1=0}
λ
1
=
i
{\displaystyle \lambda _{1}=\mathrm {i} }
λ
2
=
−
i
{\displaystyle \lambda _{2}=-\mathrm {i} }
Lauseen 5 mukaan DY:n ratkaisukannan muodostavat funktiot
x
1
(
t
)
=
cos
t
{\displaystyle x_{1}(t)=\cos t}
x
2
(
t
)
=
sin
t
{\displaystyle x_{2}(t)=\sin t}
x
3
(
t
)
=
u
1
(
t
)
x
1
(
t
)
+
u
2
(
t
)
x
2
(
t
)
=
u
1
(
t
)
cos
t
+
u
2
(
t
)
sin
t
{\displaystyle x_{3}(t)=u_{1}(t)x_{1}(t)+u_{2}(t)x_{2}(t)=u_{1}(t)\cos t+u_{2}(t)\sin t}
Ratkaistaan
u
1
{\displaystyle u_{1}}
u
2
{\displaystyle u_{2}}
W
(
x
1
,
x
2
)
(
t
)
=
det
[
x
1
(
t
)
x
2
(
t
)
x
1
′
(
t
)
x
2
′
(
t
)
]
=
det
[
cos
t
sin
t
−
sin
t
cos
t
]
=
cos
t
⋅
cos
t
−
(
−
sin
t
)
⋅
sin
t
=
cos
2
t
+
sin
2
t
=
1
∀
t
∈
]
0
,
π
2
[
{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{2}(t)\\x_{1}'(t)&x_{2}'(t)\end{matrix}}\right]\\&=\det \left[{\begin{matrix}\cos t&\sin t\\-\sin t&\cos t\end{matrix}}\right]\\&=\cos t\cdot \cos t-(-\sin t)\cdot \sin t\\&=\cos ^{2}t+\sin ^{2}t\\&=1\quad \forall ~t\in \,\left]0,{\frac {\pi }{2}}\right[\end{aligned}}}
u
1
′
(
t
)
=
−
x
2
(
t
)
q
(
t
)
W
(
x
1
,
x
2
)
(
t
)
=
−
sin
t
cos
t
=
−
tan
t
{\displaystyle u_{1}'(t)=-{\frac {x_{2}(t)q(t)}{W(x_{1},x_{2})(t)}}=-{\frac {\sin t}{\cos t}}=-\tan t}
u
2
′
(
t
)
=
x
1
(
t
)
q
(
t
)
W
(
x
1
,
x
2
)
(
t
)
=
cos
t
cos
t
=
1
{\displaystyle u_{2}'(t)={\frac {x_{1}(t)q(t)}{W(x_{1},x_{2})(t)}}={\frac {\cos t}{\cos t}}=1}
Integroidaan yhtälöt puolittain:
u
1
(
t
)
=
−
∫
tan
t
d
t
=
ln
(
cos
t
)
+
C
1
{\displaystyle u_{1}(t)=-\int \tan t\,\mathrm {d} t=\ln(\cos t)+C_{1}}
u
2
(
t
)
=
∫
1
d
t
=
t
+
C
2
{\displaystyle u_{2}(t)=\int 1\,\mathrm {d} t=t+C_{2}}
Koska
x
3
{\displaystyle x_{3}}
C
1
=
C
2
=
0
{\displaystyle C_{1}=C_{2}=0}
x
3
(
t
)
=
ln
(
cos
t
)
cos
t
+
t
sin
t
{\displaystyle x_{3}(t)=\ln(\cos t)\cos t+t\sin t}
x
:
]
0
,
π
2
[
→
R
{\displaystyle x:\,\left]0,{\frac {\pi }{2}}\right[\to \mathbb {R} }
x
(
t
)
=
c
1
cos
t
+
c
2
sin
t
+
ln
(
cos
t
)
cos
t
+
t
sin
t
{\displaystyle x(t)=c_{1}\cos t+c_{2}\sin t+\ln(\cos t)\cos t+t\sin t}
↑ 1,0 1,1 Nollafunktiota
x
(
t
)
≡
0
{\displaystyle x(t)\equiv 0}
triviaaliratkaisuksi .
↑ Epäsuorassa todistuksessa lähdetään liikkeelle alkuperäisen väitteen vastaväitteestä (eli antiteesistä). Jos antiteesin todistaminen johtaa loogiseen ristiriitaan, pitää alkuperäinen väite paikkansa.↑ Wronskin determinantti on funktio, joka määritellään kahden muun funktion avulla ja jonka arvot ovat reaalilukuja.
↑ Differentiaaliyhtälöt/Ensimmäisen kertaluvun differentiaaliyhtälöt ↑ ''Ehdot ovat yhtäpitäviä'' tarkoittaa sitä, että jos yksikin ehdoista on totta, kaikki muutkin ovat totta ja päin vastoin. Todistukseksi riittää osoittaa, että yksi ehto seuraa edellisestä, esimerkiksi järjestyksessä 1
⇒
{\displaystyle \Rightarrow }
⇒
{\displaystyle \Rightarrow }
⇒
{\displaystyle \Rightarrow }
⇒
{\displaystyle \Rightarrow }
↑ Ks. Ensimmäisen kertaluvun lineaarinen, yleinen DY
↑ Ks. Ensimmäisen kertaluvun lineaarinen, homogeeninen DY
↑ Toisen asteen yhtälön
a
t
2
+
b
t
+
c
=
0
{\displaystyle at^{2}+bt+c=0}
t
=
−
b
±
b
2
−
4
a
c
2
a
{\displaystyle t={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}
↑ Sini on pariton funktio , joten
sin
(
−
θ
)
=
−
sin
θ
{\displaystyle \sin(-\theta )=-\sin \theta }
cos
(
−
θ
)
=
cos
θ
{\displaystyle \cos(-\theta )=\cos \theta }
![{\displaystyle x:\,]x_{0}-\delta ,x_{0}+\delta [\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dcf45a6bd915779139a90b0fb4e3a3423713db0)
![{\displaystyle p,r,q:\,]a,b[\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/154ed9af80d08bc1c5f0546feaf15da86ac92cde)
![{\displaystyle {\begin{aligned}w''(t)+p(t)w'(t)+r(t)w(t)&=\alpha x_{1}''(t)+\alpha p(t)x_{1}'(t)+\alpha r(t)x_{1}(t)+\beta x_{2}''(t)+\beta p(t)x_{2}'(t)+\beta r(t)x_{2}(t)\\&=\alpha [x_{1}''(t)+p(t)x_{1}'(t)+r(t)x_{1}(t)]+\beta [x_{2}''(t)+p(t)x_{2}'(t)+r(t)x_{2}(t)]\\&=\alpha \cdot 0+\beta \cdot 0\\&=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7775b8a6bf5d09a930cf0c29f541b83089db396f)
![{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{2}(]a,b[)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/555250202b413f1bfdac66124cc71a5826a407a9)
![{\displaystyle t\in \,]a,b[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9de0fac113242e966b410763d88b3c405091ae99)
![{\displaystyle f,g:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d596877625fd914ef9a0428399115e14dc65c5c1)
![{\displaystyle x_{1},x_{2}\in {\mathcal {C}}^{1}(]a,b[)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c481d61514bd345b15c019e0ea66e7b7df6f26e8)
![{\displaystyle W(x_{1},x_{2}):\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fa07632663813d87071766d8624193aa90b361b)
![{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{1}'(t)\\x_{2}(t)&x_{2}'(t)\end{matrix}}\right]\\&=x_{1}(t)x_{2}'(t)-x_{2}(t)x_{1}'(t)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b2b8adc6b24293fe2feeb3dabd52d53b848e72a)
![{\displaystyle t_{0}\in \,]a,b[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1777be7d9968efb301f9a439a6b266a0daca3bff)
![{\displaystyle t\in \,]t_{0}-\delta ,t_{0}+\delta [}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7d0fbedb9dbb0cdf45aff64d09651fbd9791057)
![{\displaystyle {\begin{aligned}(C_{1}x_{2}(t)-C_{2}x_{1}(t))\mu (t)&=W(x,x_{1})(t)\cdot x_{2}(t)-W(x,x_{2})(t)\cdot x_{1}(t)\\&=[x(t)x_{1}'(t)-x'(t)x_{1}(t)]x_{2}(t)-[x(t)x_{2}'(t)-x'(t)x_{2}(t)]x_{1}(t)\\&=x(t)x_{1}'(t)x_{2}(t)-x(t)x_{2}'(t)x_{1}(t)-x'(t)x_{1}(t)x_{2}(t)+x'(t)x_{2}(t)x_{1}(t)\\&=-x(t)[x_{2}'(t)x_{1}(t)-x_{1}'(t)x_{2}(t)]\\&=-x(t)W(x_{1},x_{2})(t)\\&=-Cx(t)\mu (t)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ea74c283063d604506b032a157244e0426a1147)
![{\displaystyle p,r:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b3e507621cf7c0b089a1077419bdb0601aad2e9)
![{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t_{0})&=\det \left[{\begin{matrix}x_{1}(t_{0})&x_{2}(t_{0})\\x_{1}'(t_{0})&x_{2}'(t_{0})\end{matrix}}\right]\\&=\det \left[{\begin{matrix}1&1\\0&1\end{matrix}}\right]\\&=1\neq 0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7280dff3cad0c4a331b38f454977bb7df9a0895)
![{\displaystyle v:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a2abaa3cb1ca6df7372e5ce2466296c88b52935)
![{\displaystyle {\begin{aligned}0&=x_{2}''(t)+p(t)x_{2}'(t)+r(t)x_{2}(t)\\&=v''(t)x_{1}(t)+2v'(t)x_{1}'(t)+v(t)x_{1}''(t)+p(t)[v'(t)x_{1}(t)+v(t)x_{1}'(t)]+r(t)v(t)x_{1}(t)\\&=v(t)\underbrace {[x_{1}''(t)+p(t)x_{1}'+r(t)x_{1}(t)]} _{=0}+v''(t)x_{1}(t)+2v'(t)x_{1}'(t)+p(t)x_{1}(t)\\&=v''(t)x_{1}(t)+2v'(t)x'_{1}(t)+p(t)x_{1}(t)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f90b1f13c3574f7f0572f80b1020bd8b4888eb33)
![{\displaystyle s:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/008ba2dfa9d8623f422658ee7f6fa4fa8f261502)
![{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{2}(t)\\x_{1}'(t)&x_{2}'(t)\end{matrix}}\right]\\&=\det \left[{\begin{matrix}x_{1}(t)&v(t)x_{1}(t)\\x_{1}'(t)&v'(t)x_{1}(t)+v(t)x_{1}'(t)\end{matrix}}\right]\\&=x_{1}(t)[v'(t)x_{1}(t)+v(t)x_{1}'(t)]-x_{1}'(t)v(t)x_{1}(t)\\&=v'(t)x_{1}^{2}(t)\\&=z(t)x_{1}^{2}(t)\\&=\mathrm {e} ^{-\int _{t_{0}}^{t}s(t)\,\mathrm {d} s}\cdot x_{1}^{2}(t)\neq 0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f99abdbdc2f60079d5c9ac9d6084ba21c5b2c504)
![{\displaystyle {\begin{aligned}0&=x_{2}''(t)-{\frac {1}{t}}x_{2}'(t)+{\frac {1}{t^{2}}}x_{2}(t)\\&=v''(t)t+2v'(t)-{\frac {1}{t}}[v'(t)t+v(t)]+{\frac {1}{t^{2}}}v(t)t\\&=tv''(t)+2v'(t)-v'(t)-{\frac {v(t)}{t}}+{\frac {v(t)}{t}}\\&=tv''(t)+v'(t)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90def156cca8007e4c6cf52564a2f85016ba6e1d)
![{\displaystyle x:\,]0,\infty [\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb4c94e7b4e480300c46ffe22d43a37099932e61)
![{\displaystyle {\begin{aligned}0&=(1-t^{2})x_{2}''(t)-2tx_{2}'(t)+2x_{2}(t)\\&=(1-t^{2})[v''(t)t+2v'(t)]-2t[v'(t)+v(t)]+2v(t)t\\&=(1-t^{2})tv''(t)+2(1-t^{2})v'(t)-2t^{2}v'(t)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d1f208b84bd8ea9669d7caff0c0c1e3eb883a4b)
![{\displaystyle {\begin{aligned}z(t)&=\exp \left[-\int {\frac {2}{t}}\,\mathrm {d} t+\int {\frac {2t}{1-t^{2}}}\,\mathrm {d} t\right]\\&=\exp \left[-\ln t^{2}-\ln(1-t^{2})+C_{1}\right]\\&=\exp(C_{1})\cdot \exp \left[-\ln \left(t^{2}(1-t^{2})\right)\right]\\&={\frac {{\tilde {C}}_{1}}{t^{2}(1-t^{2})}},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10a98d5eaf20c5b345fe8a39d06e118eded15142)
![{\displaystyle x:\,]0,1[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/81ddaafd043448a8d77c14077361b06ad6643190)
![{\displaystyle W(x_{1},x_{2})(t)=\det \left[{\begin{matrix}\mathrm {e} ^{\lambda _{1}t}&\lambda _{1}\mathrm {e} ^{\lambda _{1}t}\\\mathrm {e} ^{\lambda _{2}t}&\lambda _{2}\mathrm {e} ^{\lambda _{2}t}\end{matrix}}\right]=(\lambda _{2}-\lambda _{1})\mathrm {e} ^{(\lambda _{1}+\lambda _{2})t}\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56b26532d77cde74ab7ee2642312e0ebf21fe093)
![{\displaystyle p,q,r:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a4bfadd6b4280a69c412986040c43bbed2b82ad)
![{\displaystyle x_{1}:\,]a,b[\,\to \mathbb {R} ,x_{1}(t)=x(t)-{\hat {x}}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40e1a1343320d7b76d65dd69a08bcc970fafd80e)
![{\displaystyle x_{1},x_{2}:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/03f20c3d938430664ea2d7b1c095ed5d9cbf3f14)
![{\displaystyle x_{3}:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/847dc59b8a10ab8afff2b2d1ed32ca63bee18e39)
![{\displaystyle x:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b113611f2a380d06ea167ac4eca169475863d3e0)
![{\displaystyle w:\,]a,b[\,\to \mathbb {R} ,w(t)=x(t)-x_{3}(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6e87e0c663a54fac6fb6337d23b4caa55e77718)
![{\displaystyle u_{1},u_{2}:\,]a,b[\,\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dce70d2d57c65e53bc12c08bfd71bbe976eb27c)
![{\displaystyle {\begin{aligned}W(x_{1},x_{2})(t)&=\det \left[{\begin{matrix}x_{1}(t)&x_{2}(t)\\x_{1}'(t)&x_{2}'(t)\end{matrix}}\right]\\&=\det \left[{\begin{matrix}\cos t&\sin t\\-\sin t&\cos t\end{matrix}}\right]\\&=\cos t\cdot \cos t-(-\sin t)\cdot \sin t\\&=\cos ^{2}t+\sin ^{2}t\\&=1\quad \forall ~t\in \,\left]0,{\frac {\pi }{2}}\right[\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e751b5f562dae9d3c04140e45ec4e7056623119)
![{\displaystyle x:\,\left]0,{\frac {\pi }{2}}\right[\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c528261d5762d89d917829a063c464d01a218a0a)
