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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - y = 3 x + cos (x)

y^{^{''}} - 2 y^{^{'}} + 10 = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 5, y^{^{'}} (0) = 3

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 3 x

(D^{2} - D) y = e^{x} sin (x)

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