What is the solution for y^{''}+4y=4t^2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=4t^2,y(0)=0,y^'(0)=0 is y=t^2+1/2 cos(2t)-1/2

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y^{''}+4y=4t^2,y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = 4 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{(- 4 x)}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{( x ^{2} + 4 )}

y^{^{''}} - y = 12 t, y (0) = - 12 t

y^{^{''}} + 4 y^{^{'}} = e^{x} cos (2 x)

(D^{2} - 6 D + 13) y = 5^{x}

What is the solution for y^{''}+4y=4t^2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=4t^2,y(0)=0,y^'(0)=0 is y=t^2+1/2 cos(2t)-1/2