What is the solution for -x^{''}-4x=t^2 ?

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

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-x^{''}-4x=t^2

- x^{^{''}} - 4 x = t^{2}

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

Solution steps

- x^{''} (t) - 4 x = t^{2}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x)

x^{^{''}} + 3 x^{^{'}} + 33 x = 9 cos (5 t)

y^{^{''}} + 4 y^{^{'}} + 3 y - 3 x = 0

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 3 y = x^{2} - 1

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 9 t

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