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

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

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(D^2+4)y=cos(2x)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 8 cos (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = (1 - x) e^{4 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 1 + x e^{x}

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

4 y^{^{''}} - y = e^{\frac{x}{2}} + 5

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