I will not tell you if instructors knew about this "money for nothing" solutions (or may be some knew but kept for themselves) but this situation is not an uncommon in the real research as well. Sometimes after a year of hard work comes an idea which solves the problem on 1--2 pages and renders useless 50 page paper already submitted. And then years after one understands that these 50 pages were relevant to another problem ...

J. Y. Yook made few mistakes uncritically applying formula (forgetting that the senior coefficient is not $1$ so that the r.h.e. must be divided by it.

\begin{equation*}

\left\{\begin{aligned}

&u'_1 t + u'_2 (t^2+1)= 0,\\

&u'_1 + u'_2 \cdot 2t= \frac{1}{t^2-1}.

\end{aligned}

\right.

\end{equation*}

Then

\begin{equation*}

\left\{\begin{aligned}

&u'_1=\frac{1}{t^2-1}-\frac{2t^2}{(t^2-1)^2},\\

&u'_2=\frac{t}{(t^2-1)^2}

\end{aligned}

\right.

\end{equation*}

and

\begin{align*}

&u_2 = \int \frac{t}{(t^2-1)^2}\,dt =-\frac{1}{2 (t^2-1)} +c_2, \text{substitution $z=t^2-1$}\\

&u_1 = \int \frac{1}{t^2-1}dt -\int \frac{2t^2}{(t^2-1)^2}dt=\int \frac{1}{t^2-1}dt + \int t\cdot d\left(\frac{1}{t^2-1}\right)= \frac{t}{t^2-1}+c_1

\end{align*}

where we integrated by parts in $u_1$. So

\begin{equation*}

y = u_1 t+ u_2 (t^2+1)= \frac{1}{2} +c_1 t+c_2 (t^2+1)

\end{equation*}

which is the general solution containing the partial solution.