What is the solution for y^{'''}-3y^{''}+3y^'-y=12e^x ?

The solution for y^{'''}-3y^{''}+3y^'-y=12e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+2e^xx^3

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y^{'''}-3y^{''}+3y^'-y=12e^x

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 12 e^{x}

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

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 3 y^{'} (x) - y = 12 e^{x}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = 8 - (t - 2)

x^{^{''}} - 16 x^{^{'}} + 64 x = 5 t e^{8 t}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 2 x + 1

0.05 y^{^{''}} + 5 y^{^{'}} + \frac{1}{0.006} y = 5 sin (25 x)

(d^{2} - d) y = e^{x} cos (x)

What is the solution for y^{'''}-3y^{''}+3y^'-y=12e^x ?
The solution for y^{'''}-3y^{''}+3y^'-y=12e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+2e^xx^3