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

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

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y^{''}+4y=xsin(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}

\frac{d}{d x} (6 + 4 x - x^{2})

\frac{d}{d x} (4 2 x - 1)

x \to 3 lim (\frac{3 x}{x - 3})

\int \frac{x + 1}{x ^{2} + 3} d x

\frac{d}{d x} (\frac{3 x ^{3} + 9 x ^{2}}{x})

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