What is the solution for y^{''}+2y^'-3y=4e^{2t} ?

The solution for y^{''}+2y^'-3y=4e^{2t} is y=c_{1}e^t+c_{2}e^{-3t}+4/5 e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'-3y=4e^{2t}

y^{^{''}} + 2 y^{^{'}} - 3 y = 4 e^{2 t}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{4}{5} e^{2 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = 4 e^{2 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{4}{5} e^{2 t}

y = c_{1} e^{t} + c_{2} e^{- 3 t} + \frac{4}{5} e^{2 t}

y^{^{''}} - y = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = 1, y (0) = 2, y^{^{'}} (0) = 1

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

y^{^{''}} + y^{^{'}} = 10 x^{4} + 2

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

What is the solution for y^{''}+2y^'-3y=4e^{2t} ?
The solution for y^{''}+2y^'-3y=4e^{2t} is y=c_{1}e^t+c_{2}e^{-3t}+4/5 e^{2t}