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

The solution for y^{''}+4y+4=x^2+2sin(x)cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)-9/8+(x^2-xcos(2x))/4

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{9}{8} + \frac{x ^{2} - x cos ( 2 x )}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{9}{8} + \frac{x ^{2} - x cos ( 2 x )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{9}{8} + \frac{x ^{2} - x cos ( 2 x )}{4}

y^{^{''}} - 2 y^{^{'}} + y = \frac{12 e ^{x}}{x ^{3}}

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 12 y = (x + 1) e^{x}

y^{^{''}} - y^{^{'}} - 2 y = cos (x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 e^{6 x}

What is the solution for y^{''}+4y+4=x^2+2sin(x)cos(x) ?
The solution for y^{''}+4y+4=x^2+2sin(x)cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)-9/8+(x^2-xcos(2x))/4