What is the solution for y^{''}+4y=sin(t) ?

The solution for y^{''}+4y=sin(t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/3 sin(t)

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y^{''}+4y=sin(t)

y^{^{''}} + 4 y = sin (t)

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

Solution steps

y^{''} (t) + 4 y = sin (t)

Solve linear ODE:

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

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

x^{- 2}, 6

y^{2} = x^{3}, y^{2} = x

\int \frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} d x

\int \frac{1}{x ^{2} x ^{2} + 49} d x

\frac{d}{d x} (\frac{2}{( 1 - \frac{1}{3} x ^{3} + 1 ) ^{2}})

What is the solution for y^{''}+4y=sin(t) ?
The solution for y^{''}+4y=sin(t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/3 sin(t)