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

The solution for y^{''}+6y^'+5y=e^x is y=c_{1}e^{-x}+c_{2}e^{-5x}+1/12 e^x

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y^{''}+6y^'+5y=e^x

y^{^{''}} + 6 y^{^{'}} + 5 y = e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{1}{12} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{1}{12} e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{1}{12} e^{x}

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

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

y^{^{'''}} - 6 y^{^{''}} + 25 y^{^{'}} = 68 e^{2 x}

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

(d - 3)^{3} y = e^{3 x}

What is the solution for y^{''}+6y^'+5y=e^x ?
The solution for y^{''}+6y^'+5y=e^x is y=c_{1}e^{-x}+c_{2}e^{-5x}+1/12 e^x