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

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

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

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

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{52} sin (2 x) - \frac{5}{52} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{52} sin (2 x) - \frac{5}{52} cos (2 x)

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{52} sin (2 x) - \frac{5}{52} cos (2 x)

y^{^{''}} + 3 y^{^{'}} + 5 y = sin (t), y^{^{'}} (0) = 0, y (0) = 1

y^{^{''}} + (4 + 1) y^{^{'}} = - 4 y + 7 e^{- x}

y^{^{''}} - y^{^{'}} + 4 y = 5 e^{2 t}

y^{^{''}} - 4 y = e^{5 x}

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 9

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