What is the solution for 3y^{''}-9y^'=sin(2x) ?

The solution for 3y^{''}-9y^'=sin(2x) is y=c_{1}e^{3x}+c_{2}-1/39 sin(2x)+1/26 cos(2x)

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3y^{''}-9y^'=sin(2x)

3 y^{^{''}} - 9 y^{^{'}} = sin (2 x)

y = c_{1} e^{3 x} + c_{2} - \frac{1}{39} sin (2 x) + \frac{1}{26} cos (2 x)

Solution steps

3 y^{''} (x) - 9 y^{'} (x) = sin (2 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{1}{39} sin (2 x) + \frac{1}{26} cos (2 x)

y = c_{1} e^{3 x} + c_{2} - \frac{1}{39} sin (2 x) + \frac{1}{26} cos (2 x)

What is the solution for 3y^{''}-9y^'=sin(2x) ?
The solution for 3y^{''}-9y^'=sin(2x) is y=c_{1}e^{3x}+c_{2}-1/39 sin(2x)+1/26 cos(2x)