What is the solution for y^{''}-3y^'+2y=12e^{-2t} ?

The solution for y^{''}-3y^'+2y=12e^{-2t} is y=c_{1}e^{2t}+c_{2}e^t+e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'+2y=12e^{-2t}

y^{^{''}} - 3 y^{^{'}} + 2 y = 12 e^{- 2 t}

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

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 12 e^{- 2 t}

Solve linear ODE:

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

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

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 90 sin (3 t)

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

y^{^{''}} - 2 y^{^{'}} + 6 = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 20

y^{^{''}} + 4 y = cos (\frac{9}{4} t)

What is the solution for y^{''}-3y^'+2y=12e^{-2t} ?
The solution for y^{''}-3y^'+2y=12e^{-2t} is y=c_{1}e^{2t}+c_{2}e^t+e^{-2t}