What is the solution for y^{''}+4y^'+5y=-5x+3e^{-x} ?

The solution for y^{''}+4y^'+5y=-5x+3e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-x+4/5+3/2 e^{-x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+5y=-5x+3e^{-x}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = - 5 x + 3 e^{- x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - x + \frac{4}{5} + \frac{3}{2} e^{- x}

What is the solution for y^{''}+4y^'+5y=-5x+3e^{-x} ?
The solution for y^{''}+4y^'+5y=-5x+3e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-x+4/5+3/2 e^{-x}