What is the solution for y^{''}+3y^'+2y=e^x ?

The solution for y^{''}+3y^'+2y=e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+1/6 e^x

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y^{''}+3y^'+2y=e^x

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = e^{x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x}

x^{^{''}} + 4 x^{^{'}} + 13 x = 5

9 y^{^{''}} - 4 y = sin (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (x)

\frac{d ^{2} y}{d x ^{2}} + y = cos (x)

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (2 t)

What is the solution for y^{''}+3y^'+2y=e^x ?
The solution for y^{''}+3y^'+2y=e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+1/6 e^x