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

The solution for y^{''}+4y=(cos(2x))/2 is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(2x))/8

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x sin ( 2 x )}{8}

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{t} + e ^{- t} )}{2}

y^{^{''}} - y = e^{- 2 t}

11 x^{^{''}} + 112.5 x = 5 e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 2 x

y^{^{''}} + 4 y = 4 cos (2 x), y^{^{'}} (\frac{π}{2}) = 0, y (0) = 1

What is the solution for y^{''}+4y=(cos(2x))/2 ?
The solution for y^{''}+4y=(cos(2x))/2 is y=c_{1}cos(2x)+c_{2}sin(2x)+(xsin(2x))/8