What is the solution for y^{''}+5y^'+4y=8x ?

The solution for y^{''}+5y^'+4y=8x is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x-5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+5y^'+4y=8x

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x

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

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 8 x

Solve linear ODE:

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

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

What is the solution for y^{''}+5y^'+4y=8x ?
The solution for y^{''}+5y^'+4y=8x is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x-5/2