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

The solution for y^{''}-y=t,y(0)=0,y^'(0)=0 is y= 1/2 e^t-1/2 e^{-t}-t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-y=t,y(0)=0,y^'(0)=0

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t} - t

Solution steps

y^{''} (t) - y = t

Solve linear ODE:

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t} - t

y = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t} - t

What is the solution for y^{''}-y=t,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y=t,y(0)=0,y^'(0)=0 is y= 1/2 e^t-1/2 e^{-t}-t