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

The solution for y^{''}+4y^'+4y=xe^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{2x}x)/(16)-1/32 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} - \frac{1}{32} e^{2 x}

2 y^{^{''}} + y^{^{'}} + 4 y = x + 1

6 y^{^{''}} - 10 y - 4 y = x + 2, y (0) = \frac{1}{8}, y^{^{'}} (0) = - \frac{1}{4}

y^{^{''}} - y^{^{'}} - 6 y = 2 x

y^{^{''}} - y^{^{'}} - 6 y = 5 - 6^{x}, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} - 2 y^{^{''}} + y = x e^{x} + e^{x}

What is the solution for y^{''}+4y^'+4y=xe^{2x} ?
The solution for y^{''}+4y^'+4y=xe^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{2x}x)/(16)-1/32 e^{2x}