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}

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