What is the solution for y^{''}+4y=1+sin(2x) ?

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

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y^{''}+4y=1+sin(2x)

y^{^{''}} + 4 y = 1 + sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = 1 + sin (2 x)

Solve linear ODE:

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

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

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