What is the solution for (d^2y)/(dx^2)+y=sin(3x)cos(2x) ?

The solution for (d^2y)/(dx^2)+y=sin(3x)cos(2x) is

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(d^2y)/(dx^2)+y=sin(3x)cos(2x)

\frac{d ^{2} y}{d x ^{2}} + y = sin (3 x) cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) + (\frac{1}{6} sin (3 x) cos (3 x) - \frac{1}{2} sin (3 x) cos (x) - \frac{1}{4} x - \frac{1}{24} sin (6 x) + \frac{3}{16} sin (4 x) + \frac{3}{8} sin (2 x)) cos (x) + (- \frac{1}{8} cos (2 x) cos (4 x) - \frac{1}{4} cos^{2} (2 x) + \frac{1}{16} cos (4 x) + \frac{1}{48} cos (6 x) - \frac{1}{16} cos (2 x)) sin (x)

Solution steps

\frac{d ^{2} y}{d x ^{2}} + y = sin (3 x) cos (2 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + (\frac{1}{6} sin (3 x) cos (3 x) - \frac{1}{2} sin (3 x) cos (x) - \frac{1}{4} x - \frac{1}{24} sin (6 x) + \frac{3}{16} sin (4 x) + \frac{3}{8} sin (2 x)) cos (x) + (- \frac{1}{8} cos (2 x) cos (4 x) - \frac{1}{4} cos^{2} (2 x) + \frac{1}{16} cos (4 x) + \frac{1}{48} cos (6 x) - \frac{1}{16} cos (2 x)) sin (x)

y = c_{1} cos (x) + c_{2} sin (x) + (\frac{1}{6} sin (3 x) cos (3 x) - \frac{1}{2} sin (3 x) cos (x) - \frac{1}{4} x - \frac{1}{24} sin (6 x) + \frac{3}{16} sin (4 x) + \frac{3}{8} sin (2 x)) cos (x) + (- \frac{1}{8} cos (2 x) cos (4 x) - \frac{1}{4} cos^{2} (2 x) + \frac{1}{16} cos (4 x) + \frac{1}{48} cos (6 x) - \frac{1}{16} cos (2 x)) sin (x)

3 y^{^{''}} + 3 y^{^{'}} - 1 = 0

5 y^{^{''}} + y^{^{'}} = - 6 x, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{'''}} - 2 y^{^{''}} + 2 y^{^{'}} = 1 + x e^{x} e

y^{^{''}} + 6 y^{^{'}} + 34 y = 68, y^{^{'}} (0) = 7, y (0) = 5

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

What is the solution for (d^2y)/(dx^2)+y=sin(3x)cos(2x) ?
The solution for (d^2y)/(dx^2)+y=sin(3x)cos(2x) is