What is the solution for y^{''}-y=e^{at} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=e^{at}

y^{^{''}} - y = e^{a t}

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

Solution steps

y^{''} (t) - y = e^{a t}

Solve linear ODE:

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

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

2 y^{^{''}} - 6 y^{^{'}} = 5

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = e^{- x} sin (x)

y^{^{''}} + 4 y = e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 5

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

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