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

The solution for y^{''}+4y=x^2 is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4-1/8

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8}

y^{^{'}} = \frac{3 x}{y}

\frac{d}{d x} (\frac{ln ( 3 x ^{2} - 6 x )}{4 x ^{3} - 2})

\int (ln (x))^{7} d x

\frac{d}{d x} (4 sec (4 x))

\int e^{(x + 1)} d x

What is the solution for y^{''}+4y=x^2 ?
The solution for y^{''}+4y=x^2 is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4-1/8