What is the solution for y^{''}-4y^'=(t^2+1)sin(2t) ?

The solution for y^{''}-4y^'=(t^2+1)sin(2t) is

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y^{''}-4y^'=(t^2+1)sin(2t)

y^{^{''}} - 4 y^{^{'}} = (t^{2} + 1) sin (2 t)

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)

Solution steps

y^{''} (t) - 4 y^{'} (t) = (t^{2} + 1) sin (2 t)

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)

y = c_{1} e^{4 t} + c_{2} + \frac{t ^{2} cos ( 2 t )}{10} - \frac{t ^{2} sin ( 2 t )}{20} - \frac{7 t sin ( 2 t )}{50} - \frac{t cos ( 2 t )}{50} + \frac{4}{125} cos (2 t) - \frac{51}{1000} sin (2 t)

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t)

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

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 x}

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

y^{^{''}} - y^{^{'}} - 2 y = 6 x + 6 e^{- x}

What is the solution for y^{''}-4y^'=(t^2+1)sin(2t) ?
The solution for y^{''}-4y^'=(t^2+1)sin(2t) is