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

The solution for y^{'''}-3y^{''}+3y^'-y=e^x 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

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

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}

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}

y^{^{''}} - y = 8

y^{^{''}} + y = H (t) - 2 H (t - \frac{b}{2}) + H (t - b)

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

y^{^{''}} + 4 y = - 5

y^{^{''}} + 2 y^{^{'}} + 4 y = 3 cos (2 t) + 4 sin (2 t)

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