What is the solution for y^{'''}-3y^{''}+3y^'-y-e^x=0 ?

The solution for y^{'''}-3y^{''}+3y^'-y-e^x=0 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6

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y^{'''}-3y^{''}+3y^'-y-e^x=0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y - e^{x} = 0

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

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 3 y^{'} (x) - y - e^{x} = 0

Solve linear ODE:

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

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

What is the solution for y^{'''}-3y^{''}+3y^'-y-e^x=0 ?
The solution for y^{'''}-3y^{''}+3y^'-y-e^x=0 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6