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

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

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

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

y = c_{1} cos (x) + c_{2} sin (x) + 2 x^{2} - 4

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + 2 x^{2} - 4

y = c_{1} cos (x) + c_{2} sin (x) + 2 x^{2} - 4

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 e^{- 2 t} sin (2 t)

y^{^{''}} + 6 y^{^{'}} + 18 y = 180, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 40 y = 4 e^{- 2 t} cos (6 t)

(D^{2} - 10 D + 25) y = e^{5 x}

9 \frac{d ^{2} y}{d x ^{2}} + 3 y = 9 (x + 5), y (0) = 5

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