What is the solution for y^{''}+4y=sin(x)*sin(2x) ?

The solution for y^{''}+4y=sin(x)*sin(2x) is

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y^{''}+4y=sin(x)*sin(2x)

y^{^{''}} + 4 y = sin (x) \cdot sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

Solution steps

y^{''} (x) + 4 y = sin (x) \cdot sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

y^{^{'}} + 2 y^{^{''}} - 4 t e^{t} = 9, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} = cos (x), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 25 y = e^{5 x}, y (0) = 1, y^{^{'}} (0) = - 4

x^{^{''}} + 25 x = t (sin (5 t) + cos (5 t))

4 y^{^{''}} - y = e^{(\frac{x}{2})} + 5

What is the solution for y^{''}+4y=sin(x)*sin(2x) ?
The solution for y^{''}+4y=sin(x)*sin(2x) is