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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{2} = 4 x, y = 2 x - 4

\frac{d}{d x} (4 x^{2} + 3 x)

\frac{d x}{d t} = \frac{x}{t}

\frac{\partial}{\partial x} (\frac{1}{∣ x - a ∣})

x = y^{2} - 6 y, x = 5 y - y^{2}

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