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

The solution for y^{''}+y^'=-cos(2x) is y=c_{1}+c_{2}e^{-x}-1/10 sin(2x)+1/5 cos(2x)

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

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

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (2 x) + \frac{1}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (2 x) + \frac{1}{5} cos (2 x)

y = c_{1} + c_{2} e^{- x} - \frac{1}{10} sin (2 x) + \frac{1}{5} cos (2 x)

y^{^{''}} - 3 y^{^{'}} + 5 = e^{2 x} - 3 cos (3 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 x

x^{^{''}} + 9 x = sin (t), x (0) = 1, x^{^{'}} (0) = 1

(D^{2} + D) (y) = sin (x)

4 y^{^{''}} - y^{^{'}} + 6 = 0

What is the solution for y^{''}+y^'=-cos(2x) ?
The solution for y^{''}+y^'=-cos(2x) is y=c_{1}+c_{2}e^{-x}-1/10 sin(2x)+1/5 cos(2x)