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}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x}

y^{^{''}} + 4 y = sin (x), y^{^{'}} (π) = 0, y (0) = 0

y^{^{'''}} - 19 y^{^{'}} + 30 y = x^{2} e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{1 + x ^{2}}

y^{^{''}} - y^{^{'}} + 144 y = 12 sin (12 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