What is the solution for y^{''}-y^'-6y=sin(3x) ?

The solution for y^{''}-y^'-6y=sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-5/78 sin(3x)+1/78 cos(3x)

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y^{''}-y^'-6y=sin(3x)

y^{^{''}} - y^{^{'}} - 6 y = sin (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) + \frac{1}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = sin (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) + \frac{1}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) + \frac{1}{78} cos (3 x)

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

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} - 4 y = e^{2 x} - 4 x + 4

y^{^{''}} - y^{^{'}} - 2 y = 15 x + 9

y^{^{''}} - y^{^{'}} - 2 y = 15 x + 8

y^{^{''}} + 6 y^{^{'}} + 25 y = 15 cos (5 t)

What is the solution for y^{''}-y^'-6y=sin(3x) ?
The solution for y^{''}-y^'-6y=sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-5/78 sin(3x)+1/78 cos(3x)