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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - e^{x} x ln (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - e^{x} x ln (x)

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - e^{x} x ln (x)

y^{^{''}} + 5 y = 4, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} - 20 y = - 56 t + 40 t^{2}

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

y^{^{''}} + y^{^{'}} + y = e^{x}

y^{^{''}} + y^{^{'}} = e^{t}

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