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

The solution for y^{''}-2y^'+y=t*e^t is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6

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y^{''}-2y^'+y=t*e^t

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t \cdot e^{t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

y = c_{1} e^{t} + c_{2} t e^{t} + \frac{e ^{t} t ^{3}}{6}

y^{^{''}} + y^{^{'}} = x^{2} e^{x}

(D^{2} - D - 2) y = 6 e^{\frac{- t}{2}}

y^{^{''}} + y^{^{'}} = 2 sin (t), y (0) = 0, y^{^{'}} (0) = - 4

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 5 y = 5 x

What is the solution for y^{''}-2y^'+y=t*e^t ?
The solution for y^{''}-2y^'+y=t*e^t is y=c_{1}e^t+c_{2}te^t+(e^tt^3)/6