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

The solution for y^{''}+5y^'+6y=sin(x) is y=c_{1}e^{-2x}+c_{2}e^{-3x}+1/10 sin(x)-1/10 cos(x)

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

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

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{10} sin (x) - \frac{1}{10} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{10} sin (x) - \frac{1}{10} cos (x)

y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x} + \frac{1}{10} sin (x) - \frac{1}{10} cos (x)

y^{^{''}} - y^{^{'}} - 6 y = 70 cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 13 y = - 2 sin (3 t)

2 y^{^{''}} + 3 y - 2 y = 6 x^{2} + 5 x - 8

\frac{3}{2} \frac{d ^{2} y}{d x ^{2}} + \frac{1}{80} \frac{d y}{d x} = g

x^{^{''}} + 2 x^{^{'}} + 5 x = e^{t}

What is the solution for y^{''}+5y^'+6y=sin(x) ?
The solution for y^{''}+5y^'+6y=sin(x) is y=c_{1}e^{-2x}+c_{2}e^{-3x}+1/10 sin(x)-1/10 cos(x)