What is the solution for y^{''}+4x^2+4y=-6 ?

The solution for y^{''}+4x^2+4y=-6 is y=c_{1}cos(2x)+c_{2}sin(2x)-x^2-1

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y^{''}+4x^2+4y=-6

y^{^{''}} + 4 x^{2} + 4 y = - 6

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

Solution steps

y^{''} (x) + 4 x^{2} + 4 y = - 6

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 1 = 2 x^{2} + x

y^{^{''}} - 4 y = 24 cos (2 t)

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 cos (x) + cos (3 x)

y^{^{''}} + 100 y^{^{'}} + 2500 y = 20

y^{^{''}} - y^{^{'}} - 2 y = 4

What is the solution for y^{''}+4x^2+4y=-6 ?
The solution for y^{''}+4x^2+4y=-6 is y=c_{1}cos(2x)+c_{2}sin(2x)-x^2-1