What is the solution for y^{''}+y=-8xcos^2(x) ?

The solution for y^{''}+y=-8xcos^2(x) is

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y^{''}+y=-8xcos^2(x)

y^{^{''}} + y = - 8 x cos^{2} (x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{8}{3} x cos^{4} (x) + \frac{8}{3} x sin^{4} (x) - 8 x sin^{2} (x) - \frac{8}{3} sin (x) cos (x) - \frac{8}{9} sin^{3} (x) cos (x) - \frac{8}{9} cos^{3} (x) sin (x)

Solution steps

y^{''} (x) + y = - 8 x cos^{2} (x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{8}{3} x cos^{4} (x) + \frac{8}{3} x sin^{4} (x) - 8 x sin^{2} (x) - \frac{8}{3} sin (x) cos (x) - \frac{8}{9} sin^{3} (x) cos (x) - \frac{8}{9} cos^{3} (x) sin (x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{8}{3} x cos^{4} (x) + \frac{8}{3} x sin^{4} (x) - 8 x sin^{2} (x) - \frac{8}{3} sin (x) cos (x) - \frac{8}{9} sin^{3} (x) cos (x) - \frac{8}{9} cos^{3} (x) sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = - 5 t e^{- t} cos (2 t)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 18 x^{2} - 4 x - 15

x^{^{''}} + x^{^{'}} = sin (t)

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

y^{^{'''}} - 3 y^{^{'}} + 2 y = 8 e^{2 x}

What is the solution for y^{''}+y=-8xcos^2(x) ?
The solution for y^{''}+y=-8xcos^2(x) is