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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{3} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{3} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{3} e^{- x}

4 y^{^{''}} - 4 y^{^{'}} + y = 40 e^{\frac{t}{2}}

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

y^{^{''}} - y^{^{'}} + 225 y = 15 sin (15 t)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 sin (x)

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