Solution:
Define $L[y] = t^2y'' - t(t+2)y' + (t+2)y$
Differentiate $y_1(t)=t$ we get $y'_1(t)=1$ and $y''_1(t)=0$.
Plug in to the given DE we see that
$$L[y_1] = t^2\cdot(0) - t(t+2)\cdot(1) + (t+2)\cdot(t) = -t^2 -2t+t^2+2t = 0$$
This shows that $y_1$ is a solution to the corresponding homogeneous equation.
Differentiate $y_2(t)=te^t$ we get $y'_2(t)=(1+t)e^t$ and $y''_2(t)=(2+t)e^t$.
Plug in to the given DE we see that
$$L[y_2] = t^2\cdot(2+t)e^t - t(t+2)\cdot(1+t)e^t + (t+2)\cdot te^t$$ $$= 2t^2e^t + t^3 e^t - (t^3 + 3t^2 + 2t)e^t + t^2 e^t + 2t e^t = 0$$
This shows that $y_2$ is a solution to the corresponding homogeneous equation.
First, let's compute the $W(y_1,y_2)$:
$$W(y_1,y_2) = det \begin{bmatrix}t&te^t\\1& (1+t)e^t\end{bmatrix} = t^2e^t$$
Now consider $Y(t) = u_1t + u_2te^t$ where $u_1$ and $u_2$ are functions of $t$.
Assume $u'_1t + u'_2te^t = 0$ and plug in $Y'$ and $Y''$ into the give DE we can get the following linear system:
$$\begin{bmatrix}t&te^t\\1& (1+t)e^t\end{bmatrix} \begin{bmatrix}u'_1\\u'_2\end {bmatrix} = \begin{bmatrix}0\\\frac{2t^3}{t^2}\end {bmatrix}$$
Then,
$$\begin{bmatrix}u'_1\\u'_2\end {bmatrix} = \frac{1}{t^2e^t} \begin{bmatrix}(1+t)e^t&-te^t\\-1&t\end{bmatrix} \begin{bmatrix}0\\2t\end {bmatrix} = \begin{bmatrix}-2\\2e^{-t}\end{bmatrix}$$
Integrate $u_1$ and $u_2$ we get:
$$u_1 = \int{-2 dt} = -2t$$ $$u_2 = \int{2e^{-t} dt} = -2e^{-t}$$
Then $Y(t) = -2t(t) + (-2e^{-t})(te^t) = -2t^2 -2t$ is a particular solution to the given nonhomogeneous equation.
Notice that $L[-2t] = 0$, thus $Y(t) = -2t^2$ is also a particular solution to the given nonhomogeneous equation.
