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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 7 x^{^{'}} + 10 x = 20, x (0) = 5, x^{^{'}} (0) = 3

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

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

\frac{d ^{2} x}{d t ^{2}} + 16 x = 1 + 2 sin (4 t)

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

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