Find a solution to the differential equation: Which is NOT a constant multiple of the solution given. Find a solution to the equation:
$$
ty'' - (t+1)y' + y =0
$$
Which is not a constant multiple of the solution: $y(t) = e^t$.

Hi, I have worked this problem and have solved it, using the particular solution of $e^t$. Why is this problem asking for a solution that is not a constant multiple of the only given solution? I do not see another way of solving it
Thank you!

(solution from comment)
Using reduction of order we aim for finding $y_2$ based on the solution given. In this case $y(t) = e^t$ is a solution. Following that we have $y_2 = g(t)  y_1$. $G$ is an unknown and will eventually be substituted for $u = g'$. Following this we differentiate $y_2$ to get $$y_2' = g_1'  e^t + g_1  e^t.$$ We differentiate again for $$y_2'' = g_1''  e^t + 2g_2'  e^t + g_2  e^t.$$ Next, we substitute $y_2$, $y_2'$ and $y_2''$ into the given equation and also substitute $u=g'$ to get $$tu' +2ut -tu -u =0.$$ Solve this equation by separation and then integrate and get the solution of y_2. We find the last $y_2 = t+1$. 

Write $$ t(y''-y') = y'-y$$
Now let $z=y'-y$, then $${z'\over z} = {1\over t}$$
so $$ \ln z = \ln t+ c\implies z = a\cdot t$$ 
for some real $a$. Now you can try if $y$ can be linear $y=kt+n$:
$$ k-kt-n = at \;\;\; \implies \;\;\; k=-a\;\;\wedge \;\;n=k=-a$$
so $y= -at-a$ is solution for every real $a$.
