What is the solution for x^{''}-2x^'+x=1t^2e^t ?

The solution for x^{''}-2x^'+x=1t^2e^t is x=c_{1}e^t+c_{2}te^t+(e^tt^4)/(12)

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x^{''}-2x^'+x=1t^2e^t

x^{^{''}} - 2 x^{^{'}} + x = 1 t^{2} e^{t}

x = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{4}}{12}

Solution steps

x^{''} (t) - 2 x^{'} (t) + x = 1 t^{2} e^{t}

Solve linear ODE:

x = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{4}}{12}

x = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{4}}{12}

What is the solution for x^{''}-2x^'+x=1t^2e^t ?
The solution for x^{''}-2x^'+x=1t^2e^t is x=c_{1}e^t+c_{2}te^t+(e^tt^4)/(12)