What is the solution for y^{''}+2y^'-3y=e^t+e^{-t} ?

The solution for y^{''}+2y^'-3y=e^t+e^{-t} is y=c_{1}e^t+c_{2}e^{-3t}+(e^tt-e^{-t})/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-3y=e^t+e^{-t}

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{t} + e^{- t}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t - e ^{- t}}{4}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = e^{t} + e^{- t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t - e ^{- t}}{4}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{e ^{t} t - e ^{- t}}{4}

\int_{0}^{64} \frac{1}{64 - x} d x

x, 4 x, - x + 22

\frac{d}{d x} (c x ln (x))

(- x^{2} sin (x) + 2 π x sin (x) - 2 π cos (x) + 2 π)^{^{'}}

\frac{d}{d x} (\frac{1 + cos ( x )}{2})

What is the solution for y^{''}+2y^'-3y=e^t+e^{-t} ?
The solution for y^{''}+2y^'-3y=e^t+e^{-t} is y=c_{1}e^t+c_{2}e^{-3t}+(e^tt-e^{-t})/4