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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=te^t

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + y = 10 sin (t)

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 3

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