What is the solution for y^{''}+y=sin(at) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=sin(at)

y^{^{''}} + y = sin (a t)

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

Solution steps

y^{''} (t) + y = sin (a t)

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 13 y = 24 e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e^{- t}

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

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

y^{^{''}} + 5 y^{^{'}} + 4 y = t e^{- t} H (t)

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