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

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

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

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

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

Solution steps

y^{''} (x) + 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}

\int x^{2} arctan (3 x) d x

\frac{\partial}{\partial x} (4 x + 5)

f (x) = x^{2} - 3

x^{2} y^{^{'}} = y - x y

\frac{d}{d x} (2 (sin (x^{2}) + 2 x^{2} cos (x^{2})))

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