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

The solution for y^{''}-3y^'=sin(2x) is y=c_{1}e^{3x}+c_{2}-1/13 sin(2x)+3/26 cos(2x)

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

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

y = c_{1} e^{3 x} + c_{2} - \frac{1}{13} sin (2 x) + \frac{3}{26} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{1}{13} sin (2 x) + \frac{3}{26} cos (2 x)

y = c_{1} e^{3 x} + c_{2} - \frac{1}{13} sin (2 x) + \frac{3}{26} cos (2 x)

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 1

y^{^{''}} + 4 y = 9 sin (t), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + y^{^{'}} = 2 e^{x}

9 y^{^{''}} - 16 y = sin (x)

What is the solution for y^{''}-3y^'=sin(2x) ?
The solution for y^{''}-3y^'=sin(2x) is y=c_{1}e^{3x}+c_{2}-1/13 sin(2x)+3/26 cos(2x)