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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} - 4) y = 4 x - 3 e^{x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + 1 = 0

y^{^{''}} + 9 y = sin (2 x), y (0) = - 1, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + y = e^{3} + e^{2 x}

y^{^{''}} + y^{^{'}} + 12 y = 20 cos (3 t)

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