What is the solution for y^{''}+5y^'+4y=2xe^{2x} ?

The solution for y^{''}+5y^'+4y=2xe^{2x} is y=c_{1}e^{-x}+c_{2}e^{-4x}+(e^{2x}x)/9-1/18 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+5y^'+4y=2xe^{2x}

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 x e^{2 x}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{9} - \frac{1}{18} e^{2 x}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 2 x e^{2 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{9} - \frac{1}{18} e^{2 x}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{9} - \frac{1}{18} e^{2 x}

What is the solution for y^{''}+5y^'+4y=2xe^{2x} ?
The solution for y^{''}+5y^'+4y=2xe^{2x} is y=c_{1}e^{-x}+c_{2}e^{-4x}+(e^{2x}x)/9-1/18 e^{2x}