What is the solution for y^{''}+y-cos(2x)=0 ?

The solution for y^{''}+y-cos(2x)=0 is y=c_{1}cos(x)+c_{2}sin(x)-1/3 cos(2x)

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y^{''}+y-cos(2x)=0

y^{^{''}} + y - cos (2 x) = 0

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

Solution steps

y^{''} (x) + y - cos (2 x) = 0

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

y^{^{'''}} - 2 y^{^{''}} + 10 y^{^{'}} = 250 e^{5 x}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 = 0

y^{^{''}} - 9 y = 4 e^{- 5 t}

y^{^{''}} + 4 y = 3 cos (a t)

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 7 e^{- 3 x}

What is the solution for y^{''}+y-cos(2x)=0 ?
The solution for y^{''}+y-cos(2x)=0 is y=c_{1}cos(x)+c_{2}sin(x)-1/3 cos(2x)