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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = x - 2 - 2 cos (8)

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

(d^{2} - 4 d + 3) y = e^{- 2 t}

y^{^{''}} + y^{^{'}} - 12 = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 1296 t e^{- t}

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