What is the solution for -y^{''}-y+x^2=0 ?

The solution for -y^{''}-y+x^2=0 is y=c_{1}cos(x)+c_{2}sin(x)+x^2-2

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-y^{''}-y+x^2=0

- y^{^{''}} - y + x^{2} = 0

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

Solution steps

- y^{''} (x) - y + x^{2} = 0

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

x \to \infty lim (\frac{3 x}{2})

\int \frac{x ^{2} - x + 18}{x ^{3} + 3 x} d x

\int \frac{1}{10 x - x ^{2}} d x

\frac{d}{d x} (0.2 e^{5 x})

f (x) = sin^{5} (3 x^{2} - x + 13)

What is the solution for -y^{''}-y+x^2=0 ?
The solution for -y^{''}-y+x^2=0 is y=c_{1}cos(x)+c_{2}sin(x)+x^2-2