What is the solution for y^{''}+y=sin(3x) ?

The solution for y^{''}+y=sin(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/8 sin(3x)

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y^{''}+y=sin(3x)

y^{^{''}} + y = sin (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{8} sin (3 x)

Solution steps

y^{''} (x) + y = sin (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{8} sin (3 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{8} sin (3 x)

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 16 e^{- 2 t}

y^{^{''}} - 2 y^{^{'}} = cos (4 x)

4 y^{^{''}} - y = 12 + 6 x

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2 t}

y^{^{''}} - 2 y^{^{'}} + 3 y = e^{x}

What is the solution for y^{''}+y=sin(3x) ?
The solution for y^{''}+y=sin(3x) is y=c_{1}cos(x)+c_{2}sin(x)-1/8 sin(3x)