What is the solution for y^{''}+y^'=-16cos(3x) ?

The solution for y^{''}+y^'=-16cos(3x) is y=c_{1}+c_{2}e^{-x}-8/15 sin(3x)+8/5 cos(3x)

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y^{''}+y^'=-16cos(3x)

y^{^{''}} + y^{^{'}} = - 16 cos (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{8}{15} sin (3 x) + \frac{8}{5} cos (3 x)

Solution steps

y^{''} (x) + y^{'} (x) = - 16 cos (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - \frac{8}{15} sin (3 x) + \frac{8}{5} cos (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{8}{15} sin (3 x) + \frac{8}{5} cos (3 x)

1 y^{^{''}} + 5 y^{^{'}} + \frac{1}{\frac{1}{6}} y = 2 e^{2 t}

y^{^{''}} + 2 y^{^{'}} - 2 y = 5 sin (x) - 10 x e^{x}

y^{^{''}} - 8 y^{^{'}} + 17 y = - 102, y (0) = - 3, y^{^{'}} (0) = 8

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x ln (x)

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 = 0

What is the solution for y^{''}+y^'=-16cos(3x) ?
The solution for y^{''}+y^'=-16cos(3x) is y=c_{1}+c_{2}e^{-x}-8/15 sin(3x)+8/5 cos(3x)