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)

y^{^{''}} + 9 y = 3, y (0) = 0, y^{^{'}} (0) = 0

8 y^{^{''}} + 7 y^{^{'}} - y = 17

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{3 x} - 7 e^{5 x}

y^{^{''}} + 0.01 \cdot (20 - y) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 6 e^{5 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)