What is the solution for y^{''}+2y^'=te^{-t} ?

The solution for y^{''}+2y^'=te^{-t} is y=c_{1}+c_{2}e^{-2t}-e^{-t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'=te^{-t}

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

y = c_{1} + c_{2} e^{- 2 t} - e^{- t} t

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} - e^{- t} t

y = c_{1} + c_{2} e^{- 2 t} - e^{- t} t

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 cos (x) - 2 sin (x)

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{2} + 5

(D + 3) y = x + 1

y^{^{''}} - y = \frac{( 4 e ^{- x} )}{( 1 - e ^{- 2 x} )}

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 x + 2

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