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

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

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y^{''}+4y=-t^2

y^{^{''}} + 4 y = - t^{2}

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 9 x = - 13 cos (2 t)

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

y^{^{''}} - 4 y = sinh (2 x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y (0) = - 1

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