What is the solution for y^{''}+4y=x-1/20 x^2 ?

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

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y^{''}+4y=x-1/20 x^2

y^{^{''}} + 4 y = x - \frac{1}{20} x^{2}

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{1}{20} x^{2}

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}

t^{3} - 12 t^{2} + 36 t

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18 cos (x) + 9 sin^{2} (x)

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