What is the solution for y^{''}+2y^'+5y=e^{-t}sin(t) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+5y=e^{-t}sin(t)

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- t} sin (t)

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{3} e^{- t} sin (t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = e^{- t} sin (t)

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{3} e^{- t} sin (t)

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + \frac{1}{3} e^{- t} sin (t)

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

(d^{3} - 7 d - 6) y = e^{2} x (1 + x)

(d^{4} - 1) y = 7 x^{2}

(d^{3} - 5 d^{2} + 8 d - 4) y = e^{2} x

(d + 2 d^{2} + 1) y = cos (x)

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