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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{3}{2} e^{2 x} d x

x y^{^{'}} + y = y^{2}, y (1) = - 5

(4 e^{2 x})^{^{'}}

(x) \to (0) lim (\frac{1}{2})

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