What is the solution for y^{''}+4y=8x^2,y(0)=-3,y^'(0)=0 ?

The solution for y^{''}+4y=8x^2,y(0)=-3,y^'(0)=0 is y=2x^2-2cos(2x)-1

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y^{''}+4y=8x^2,y(0)=-3,y^'(0)=0

y^{^{''}} + 4 y = 8 x^{2}, y (0) = - 3, y^{^{'}} (0) = 0

y = 2 x^{2} - 2 cos (2 x) - 1

Solution steps

y^{''} (x) + 4 y = 8 x^{2}

Solve linear ODE:

y = 2 x^{2} - 2 cos (2 x) - 1

y = 2 x^{2} - 2 cos (2 x) - 1

What is the solution for y^{''}+4y=8x^2,y(0)=-3,y^'(0)=0 ?
The solution for y^{''}+4y=8x^2,y(0)=-3,y^'(0)=0 is y=2x^2-2cos(2x)-1