What is the solution for y^{''}+y=2tcos(t)cos^2(t) ?

The solution for y^{''}+y=2tcos(t)cos^2(t) is

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y^{''}+y=2tcos(t)cos^2(t)

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

y = c_{1} cos (t) + c_{2} sin (t) + \frac{24 t ^{2} sin ( t ) + 32 t cos ^{5} ( t ) - 12 t cos ( t ) + 128 t cos ^{3} ( t ) sin ^{2} ( t ) - 12 t sin ( t ) sin ( 4 t ) + 3 cos ( t ) sin ( 4 t ) - 3 sin ( t ) cos ( 4 t )}{64}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + \frac{24 t ^{2} sin ( t ) + 32 t cos ^{5} ( t ) - 12 t cos ( t ) + 128 t cos ^{3} ( t ) sin ^{2} ( t ) - 12 t sin ( t ) sin ( 4 t ) + 3 cos ( t ) sin ( 4 t ) - 3 sin ( t ) cos ( 4 t )}{64}

y = c_{1} cos (t) + c_{2} sin (t) + \frac{24 t ^{2} sin ( t ) + 32 t cos ^{5} ( t ) - 12 t cos ( t ) + 128 t cos ^{3} ( t ) sin ^{2} ( t ) - 12 t sin ( t ) sin ( 4 t ) + 3 cos ( t ) sin ( 4 t ) - 3 sin ( t ) cos ( 4 t )}{64}

y^{^{''}} - 2 y^{^{'}} - 8 y = 5 e^{2} x

(d^{2} + 1) y = e^{- x} + cos (x) + x^{3}

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

(2 d^{2} - d - 3) y = 25 e^{- x}

(d^{2} - 4 d + 4) \cdot y = 8 x^{2} + e^{2 x}

What is the solution for y^{''}+y=2tcos(t)cos^2(t) ?
The solution for y^{''}+y=2tcos(t)cos^2(t) is