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

The solution for y^{''}-3y^'=sin(3x) is y=c_{1}e^{3x}+c_{2}-1/18 sin(3x)+1/18 cos(3x)

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

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

y = c_{1} e^{3 x} + c_{2} - \frac{1}{18} sin (3 x) + \frac{1}{18} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{1}{18} sin (3 x) + \frac{1}{18} cos (3 x)

y = c_{1} e^{3 x} + c_{2} - \frac{1}{18} sin (3 x) + \frac{1}{18} cos (3 x)

x^{^{''}} - c^{2} x = c cos (c t + a)

x^{^{''}} - 3 x^{^{'}} + 2 x = \frac{( e ^{2 t} )}{e ^{2 t} + 1}

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

y^{^{''}} - y^{^{'}} - 2 y = e^{- 2 t} - 3 e^{- t}

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

What is the solution for y^{''}-3y^'=sin(3x) ?
The solution for y^{''}-3y^'=sin(3x) is y=c_{1}e^{3x}+c_{2}-1/18 sin(3x)+1/18 cos(3x)