What is the solution for y^{'''}+3y^{''}-4y=e^{-2x} ?

The solution for y^{'''}+3y^{''}-4y=e^{-2x} is y=c_{1}e^x+c_{2}e^{-2x}+c_{3}xe^{-2x}-(e^{-2x}x^2)/6

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y^{'''}+3y^{''}-4y=e^{-2x}

y^{^{'''}} + 3 y^{^{''}} - 4 y = e^{- 2 x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + c_{3} x e^{- 2 x} - \frac{e ^{- 2 x} x ^{2}}{6}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + c_{3} x e^{- 2 x} - \frac{e ^{- 2 x} x ^{2}}{6}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + c_{3} x e^{- 2 x} - \frac{e ^{- 2 x} x ^{2}}{6}

y^{^{''}} + y^{^{'}} - 2 y + 1 = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 = 3 cos (x)

y^{^{''}} + 4 y = - 1

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

(d^{2} - 2 d + 2) y = 4 x - 2 + 2 e^{x} sin (x)

What is the solution for y^{'''}+3y^{''}-4y=e^{-2x} ?
The solution for y^{'''}+3y^{''}-4y=e^{-2x} is y=c_{1}e^x+c_{2}e^{-2x}+c_{3}xe^{-2x}-(e^{-2x}x^2)/6