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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{x} - 9

5 y^{^{''}} - 10 y^{^{'}} + 10 y = e^{x} sec (x)

y^{^{''}} - 16 y = 32 + 45 e^{x}

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