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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - y = sin (x) + e^{- 5 x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 1

y^{^{''}} + 9 y = 2 cos (3 x) + 2 sin (3 x)

y^{^{''}} - 32 y = 16 cos (x)

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