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

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

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

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

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

Solution steps

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

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

4 y^{^{''}} - 3 y^{^{'}} = 4 x - 6

y^{^{''}} - 9 y^{^{'}} + 8 y = x, y (0) = e^{x}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 t e^{2 t} (t + 1)

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 e^{4 x}, y (0) = 0, y^{^{'}} (0) = 1

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