What is the solution for y^{''}+4y^'+4y=t+e^{2t} ?

The solution for y^{''}+4y^'+4y=t+e^{2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(t-1)/4+1/16 e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=t+e^{2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = t + e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t - 1}{4} + \frac{1}{16} e^{2 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = t + e^{2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t - 1}{4} + \frac{1}{16} e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t - 1}{4} + \frac{1}{16} e^{2 t}

f (x) = 9 x^{2}

\int - 2 x^{3} + 12 x^{2} - 20 x + 8.5 d x

\frac{d y}{d x} + 7 x e^{y} = 0

\frac{d}{d x} (lo g_{4} (sec (x)))

\frac{d}{d x} (3 x^{2} - \frac{1}{x ^{2}})

What is the solution for y^{''}+4y^'+4y=t+e^{2t} ?
The solution for y^{''}+4y^'+4y=t+e^{2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+(t-1)/4+1/16 e^{2t}