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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

11 x^{^{''}} + 112.5 x = 5 e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 2 x

y^{^{''}} + 4 y = 4 cos (2 x), y^{^{'}} (\frac{π}{2}) = 0, y (0) = 1

x^{^{''}} + 9 x = sin (t)

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

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