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

The solution for y^{'''}-3y^{''}+3y^'-y=0 is y=c_{1}e^t+c_{2}te^t+c_{3}t^2e^t

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

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

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

Solution steps

y^{'''} (t) - 3 y^{''} (t) + 3 y^{'} (t) - y = 0

Solve linear ODE:

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

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

x \to 4 + lim (\frac{2}{x ^{2} - 16})

x \to \infty lim (x^{3 x})

y^{^{'}} - \frac{y}{x} = - x

y = \frac{ln ( e ^{x} )}{x}

\frac{d}{d x} (3 x^{2} ln (x))

What is the solution for y^{'''}-3y^{''}+3y^'-y=0 ?
The solution for y^{'''}-3y^{''}+3y^'-y=0 is y=c_{1}e^t+c_{2}te^t+c_{3}t^2e^t