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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=t,y^'(0)=1

y^{^{''}} - y = t, y^{^{'}} (0) = 1

y = (c_{2} + 2) e^{t} + c_{2} e^{- t} - t

Solution steps

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

Solve linear ODE:

y = (c_{2} + 2) e^{t} + c_{2} e^{- t} - t

y = (c_{2} + 2) e^{t} + c_{2} e^{- t} - t

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

2 y^{^{''}} + y^{^{'}} = cos^{2} (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = (2 x - 1) e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{2 e ^{x}}{x ^{2} - 1}

y^{^{''}} + 5 y^{^{'}} + 6 y = - 19 t e^{(2 t)}

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