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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + y = 3 x + 8

y^{^{''}} + 14 y^{^{'}} + 49 y = 4 t e^{- 7 t}

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{x}{2}}

y^{^{''}} - y^{^{'}} = e^{x} sin (x)

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

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