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 $y_2 = t+1$. 
 A: 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 $a$.
A: $$ty'' - (t+1)y' + y =0$$
$$ty'' - ty'-1y' + y =0$$
$$ty'' -y'-( ty'-y) =0$$
Divide by the integrating factor $\mu(t)=t^2$
$$\frac {ty'' -y'}{t^2}-\frac {( ty'-y)}{t^2} =0$$
$$(\frac {y'}t)'-(\frac yt)' =0$$
Integrate
$$\frac {y'}t-\frac yt =K_1$$
$$(ye^{-t})'=K_1te^{-t}$$
Integrate again
$$ye^{-t}=-K_1(t+1)e^{-t}+K_2$$
$$\implies y(t)=C_1(t+1)+C_2e^t$$
A: Well, one way we could go is by power series. Letting $y=\sum_{k=0}^\infty a_kt^k,$ we have: $$\begin{eqnarray}ty''&=&t\sum_{k=2}^\infty k(k-1)a_kt^{k-2}\\&=&\sum_{k=1}^\infty(k+1)ka_{k+1}t^k\end{eqnarray}\tag{1}$$ $$\begin{eqnarray}(t+1)y'&=&(t+1)\sum_{k=1}^\infty ka_kt^{k-1}\\&=&\sum_{k=1}^\infty ka_kt^k+\sum_{k=1}^\infty ka_kt^{k-1}\\&=&a_1+\sum_{k=1}^\infty ka_kt^k+\sum_{k=2}^\infty ka_kt^{k-1}\\&=&a_1+\sum_{k=1}^\infty \left(ka_k+(k+1)a_{k+1}\right)t^k\end{eqnarray}\tag{2}$$
Putting this into the given equation, we have $$0=a_0-a_1+\sum_{k=1}^\infty\left((k+1)ka_{k+1}-ka_k-(k+1)a_{k+1}+a_k\right)t^k,$$ or $$0=a_0-a_1+\sum_{k=2}^\infty(k-1)\left((k+1)a_{k+1}-a_k\right)t^k.$$ Consequently, we must have $a_0-a_1=0,$ and for all $k\ge 2,$ we must have $$(k+1)a_{k+1}-a_k=0.$$ Put another way, we must have $a_1=a_0,$ and for $k\ge 2,$ we must have $a_{k+1}=\frac1{k+1}a_k.$ Thus, all of the coefficients are completely determined by the values of $a_0$ and $a_2,$ since by an easy inductive argument, we have for all $k\ge2$ that $$a_k=\frac2{k!}a_2.$$
Thus, we have $$y=\sum_{k=0}^\infty a_kt^k=a_0(1+t)+2a_2\sum_{k=2}^\infty\frac1{k!}t^k,$$ and so $$y=a_0(1+t)+2a_2\left(e^t-1-t\right)=(a_0-2a_2)(1+t)+2a_2e^t=c_1(1+t)+c_2e^t.$$ Thus, a particular solution that is not a constant multiple of $e^t$ is (for example) $t+1.$
