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^xx)/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}

\int \frac{3 y}{x}

\frac{d}{d x} (cos (x) - 1 + \frac{1}{2} x^{2})

f (x) = \frac{1}{4 x ^{2} + 2 x}

(x^{2} + 2) (x - 5)

\frac{\partial}{\partial x} ((x + 1)^{2} + y^{2} - 1)

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^xx)/9