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}

y^{^{''}} - 4 y^{^{'}} + 4 y = 8 x^{2} e^{2 x} sin (2 x)

y^{^{''}} - y^{^{'}} = x [e^{x}], y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} - 8 y = - 2 e^{2 x} - 5 x e^{- 4 x} - 2 x

D, D (t) = 2000 sin (\frac{π t}{90}) + 4000

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 16

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)