What is the solution for y^{''}+25y=e^{-t} ?

The solution for y^{''}+25y=e^{-t} is y=c_{1}cos(5t)+c_{2}sin(5t)+1/26 e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

Solution steps

y^{''} (t) + 25 y = e^{- t}

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{26} e^{- t}

y^{^{''}} - y = e^{x} cos (x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 6 y = e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 6} e^{5 t}

y^{^{''}} - y^{^{'}} - 2 y = x e^{4 x}

q^{^{''}} + 100 q^{^{'}} + 2500 q = 40, q (0) = 0

What is the solution for y^{''}+25y=e^{-t} ?
The solution for y^{''}+25y=e^{-t} is y=c_{1}cos(5t)+c_{2}sin(5t)+1/26 e^{-t}