What is the solution for y^{''}+4y=x-(x^2)/(20) ?

The solution for y^{''}+4y=x-(x^2)/(20) is y=c_{1}cos(2x)+c_{2}sin(2x)-(x^2)/(80)+x/4+1/160

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y^{''}+4y=x-(x^2)/(20)

y^{^{''}} + 4 y = x - \frac{x ^{2}}{20}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2}}{80} + \frac{x}{4} + \frac{1}{160}

Solution steps

y^{''} (x) + 4 y = x - \frac{x ^{2}}{20}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2}}{80} + \frac{x}{4} + \frac{1}{160}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2}}{80} + \frac{x}{4} + \frac{1}{160}

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

2 π x^{2}

y^{^{'}} + \frac{1}{200 + t} y = 0

x \to 1 + lim (x^{2} - 1)

\frac{d}{d x} (\frac{a + b t + c t ^{2}}{t})

What is the solution for y^{''}+4y=x-(x^2)/(20) ?
The solution for y^{''}+4y=x-(x^2)/(20) is y=c_{1}cos(2x)+c_{2}sin(2x)-(x^2)/(80)+x/4+1/160