What is the solution for y^{''}-2y^'+y=e^t,y(0)=0,y(0)=5 ?

The solution for y^{''}-2y^'+y=e^t,y(0)=0,y(0)=5 is y=(e^tt^2)/2+c_{2}e^tt

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=e^t,y(0)=0,y(0)=5

y^{^{''}} - 2 y^{^{'}} + y = e^{t}, y (0) = 0, y (0) = 5

y = \frac{e ^{t} t ^{2}}{2} + c_{2} e^{t} t

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = e^{t}

Solve linear ODE:

y = \frac{e ^{t} t ^{2}}{2} + c_{2} e^{t} t

y = \frac{e ^{t} t ^{2}}{2} + c_{2} e^{t} t

y^{^{''}} + 16 y = 3 sec (4 x)

y^{^{''}} + 4 y = 6 cos (x)

y^{^{''}} + 9 y^{^{'}} = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x - sin (2 x)

y^{^{''}} - y = 3 x e^{x} cos (2 x)

What is the solution for y^{''}-2y^'+y=e^t,y(0)=0,y(0)=5 ?
The solution for y^{''}-2y^'+y=e^t,y(0)=0,y(0)=5 is y=(e^tt^2)/2+c_{2}e^tt