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

The solution for y^{''}+y^'=3sin(3x) is y=c_{1}+c_{2}e^{-x}-3/10 sin(3x)-1/10 cos(3x)

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y^{''}+y^'=3sin(3x)

y^{^{''}} + y^{^{'}} = 3 sin (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{3}{10} sin (3 x) - \frac{1}{10} cos (3 x)

Solution steps

y^{''} (x) + y^{'} (x) = 3 sin (3 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - \frac{3}{10} sin (3 x) - \frac{1}{10} cos (3 x)

y = c_{1} + c_{2} e^{- x} - \frac{3}{10} sin (3 x) - \frac{1}{10} cos (3 x)

y^{^{''}} - 2 y = 18 (cot (t))^{3}

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

- 4 y^{^{''}} - 4 y - y = - (t^{2} + t + 6 e^{- 2} t)

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

y^{^{''}} + 5 y^{^{'}} + 6 = sin (x) - cos (2 x)

What is the solution for y^{''}+y^'=3sin(3x) ?
The solution for y^{''}+y^'=3sin(3x) is y=c_{1}+c_{2}e^{-x}-3/10 sin(3x)-1/10 cos(3x)