What is the solution for y^{''}+2y^'+5y=e^{-x} ?

The solution for y^{''}+2y^'+5y=e^{-x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/4 e^{-x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{- x}

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{- x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{4} e^{- x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x^{2} - 8 x

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} = x

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

What is the solution for y^{''}+2y^'+5y=e^{-x} ?
The solution for y^{''}+2y^'+5y=e^{-x} is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/4 e^{-x}