What is the solution for y^{''}+4y^'+5y=35e-4x ?

The solution for y^{''}+4y^'+5y=35e-4x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-(4x)/5+(175e+16)/(25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+5y=35e-4x

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{4 x}{5} + \frac{175 e + 16}{25}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 35 e - 4 x

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{4 x}{5} + \frac{175 e + 16}{25}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) - \frac{4 x}{5} + \frac{175 e + 16}{25}

What is the solution for y^{''}+4y^'+5y=35e-4x ?
The solution for y^{''}+4y^'+5y=35e-4x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-(4x)/5+(175e+16)/(25)