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^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 2

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + 3 e^{- x}

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 + e^{- t}

y^{^{''}} + 16 y = e^{- 4 t} + 3 sin (4 t)

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