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}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 6 sin (x))

y^{^{''}} + 49 y = 13 sec^{2} (7 t)

y^{^{''}} - 10 y^{^{'}} + 25 y = 4 - e^{t}

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 2

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}