równanie
Eleqdi: Rozwiąż równanie różniczkowe x''(t)+tx'(t)+t2=0, t∊(0,1)
3 lip 10:27
Mariusz:
Równanie liniowe niejednorodne drugiego rzędu , sprowadzalne do równania pierwszego rzędu
x''(t)+tx'(t)+t
2=0
x''(t)+tx'(t)+t
2=0
y=x'(t)
y'+ty+t
2=0
y'+ty=0
y'=−ty
|y|=e
C1e
−t2/2
y=±e
C1e
−t2/2
y=C
2e
−t2/2
y=C(t)e
−t2/2
C'(t)e
−t2/2−te
−t2/2C(t)+tC(t)e
−t2/2+t
2=0
C'(t)e
−t2/2=−t
2
C'(t)=−t
2e
t2/2
C(t)=−te
t2/2+∫e
t2/2dx
∫e
t2/2dt
| t | | 1 | | t2 | |
C(t)=−tet2/2− |
| Γ( |
| ,− |
| ) |
| √−2t2 | | 2 | | 2 | |
| t | | 1 | | t2 | |
y=(−tet2/2− |
| Γ( |
| ,− |
| )+C1)e−t2/2 |
| √−2t2 | | 2 | | 2 | |
| t | | 1 | | t2 | |
y=−t− |
| Γ( |
| ,− |
| )e−t2/2+C1e−t2/2 |
| √−2t2 | | 2 | | 2 | |
| t | | 1 | | t2 | |
x'(t)=−t− |
| Γ( |
| ,− |
| )e−t2/2+C1e−t2/2 |
| √−2t2 | | 2 | | 2 | |
| t | | 1 | | t2 | |
dx = (−t− |
| Γ( |
| ,− |
| )e−t2/2+C1e−t2/2)dt |
| √−2t2 | | 2 | | 2 | |
3 lip 14:33