What is the solution for y^{''}+2y^'+2y=xe^{-x} ?

The solution for y^{''}+2y^'+2y=xe^{-x} is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+e^{-x}x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- x} x

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- x} x

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + e^{- x} x

y^{^{'''}} - 21 y^{^{''}} + 1372 y = e^{14 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 4 e^{- 3 x}

(D^{2} - 1) y = e^{x} + 1

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y (0) = 0

y^{^{''}} + y^{^{'}} + y = sin (t)

What is the solution for y^{''}+2y^'+2y=xe^{-x} ?
The solution for y^{''}+2y^'+2y=xe^{-x} is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+e^{-x}x