What is the solution for y^{'''}-3y^{''}+4y=e^{-x} ?

The solution for y^{'''}-3y^{''}+4y=e^{-x} is y=c_{1}e^{-x}+c_{2}e^{2x}+c_{3}xe^{2x}+(e^{-x}x)/9

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y^{'''}-3y^{''}+4y=e^{-x}

y^{^{'''}} - 3 y^{^{''}} + 4 y = e^{- x}

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{- x} x}{9}

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 4 y = e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{- x} x}{9}

y = c_{1} e^{- x} + c_{2} e^{2 x} + c_{3} x e^{2 x} + \frac{e ^{- x} x}{9}

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

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = - 6 e x + 10 cos (x)

\frac{d ^{3} y}{d x ^{3}} - \frac{d ^{2} y}{d x ^{2}} = e^{x}

x^{^{''}} - 18 x^{^{'}} + 81 x = t e^{9 t}

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

What is the solution for y^{'''}-3y^{''}+4y=e^{-x} ?
The solution for y^{'''}-3y^{''}+4y=e^{-x} is y=c_{1}e^{-x}+c_{2}e^{2x}+c_{3}xe^{2x}+(e^{-x}x)/9