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

The solution for y^{''}+4y=x(sin(2x)) is y=c_{1}cos(2x)+c_{2}sin(2x)-(x^2cos(2x))/8+(xsin(2x))/(16)

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2} cos ( 2 x )}{8} + \frac{x sin ( 2 x )}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2} cos ( 2 x )}{8} + \frac{x sin ( 2 x )}{16}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x ^{2} cos ( 2 x )}{8} + \frac{x sin ( 2 x )}{16}

What is the solution for y^{''}+4y=x(sin(2x)) ?
The solution for y^{''}+4y=x(sin(2x)) is y=c_{1}cos(2x)+c_{2}sin(2x)-(x^2cos(2x))/8+(xsin(2x))/(16)