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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{x ^{2}}{( x + 2 )}

ln (x^{8})

\int \frac{sin ^{2} ( θ )}{cos ^{4} ( θ )} d θ

\frac{\partial}{\partial x} (- \frac{v ( x , t )}{b ( x , t ) ( x - v ( x , t ) t )})

\frac{d y}{y} = d x

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