What is the solution for y^{''}-5y^'+6y=-3sin(2x) ?

The solution for y^{''}-5y^'+6y=-3sin(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}-3/52 sin(2x)-15/52 cos(2x)

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y^{''}-5y^'+6y=-3sin(2x)

y^{^{''}} - 5 y^{^{'}} + 6 y = - 3 sin (2 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{3}{52} sin (2 x) - \frac{15}{52} cos (2 x)

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = - 3 sin (2 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{3}{52} sin (2 x) - \frac{15}{52} cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{3}{52} sin (2 x) - \frac{15}{52} cos (2 x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 18

y^{^{''}} + y^{^{'}} + y = - 13 sin (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 2 e^{4 t}, y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} - 5 y^{^{'}} + 6 y = sin (4 x)

What is the solution for y^{''}-5y^'+6y=-3sin(2x) ?
The solution for y^{''}-5y^'+6y=-3sin(2x) is y=c_{1}e^{3x}+c_{2}e^{2x}-3/52 sin(2x)-15/52 cos(2x)