What is the solution for y^{''}+4y=1-t^3 ?

The solution for y^{''}+4y=1-t^3 is y=c_{1}cos(2t)+c_{2}sin(2t)-(t^3)/4+(3t)/8+1/4

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y^{''}+4y=1-t^3

y^{^{''}} + 4 y = 1 - t^{3}

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

Solution steps

y^{''} (t) + 4 y = 1 - t^{3}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = sin (2 x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 1 + x - 7 e^{- 2 x}

y^{^{''}} + 4 y = cos (3 t)

y^{^{''}} + 4 y = 4 cos (x) + 3 sin (x) + 8

What is the solution for y^{''}+4y=1-t^3 ?
The solution for y^{''}+4y=1-t^3 is y=c_{1}cos(2t)+c_{2}sin(2t)-(t^3)/4+(3t)/8+1/4