Reduction of Order: $t^{2}y''+3ty'+y=0$, $\quad t>0$; $\quad y_{1}(t)=t^{-1}$ I am working an exercise from Elementary Differential Equations and Boundary Value Problems  Ninth Edition by Boyce and Diprima, and I think there is mistake\typo. On page 173 Section 3.4 exercise 25. 
The book is correct I dropped the minus sign on the first integral (as pointed out by user40615 below) fixing that leads to the correct answer, also as Artem pointed out my Mathematica code had a typo!
The question is as follows: In each of the following problems 23 through 30 us the method of reduction of order to find a second solution to the given DFQ's. 
(25). $t^{2}y''+3ty'+y=0$, $\quad t>0$;  $\quad y_{1}(t)=t^{-1}$ .
The solution in the back of the book is $y_2(t)=t^{-1}\ln t$. I used Mathematica to check the solution and it gives: 

 Which translates to $y(t)\to c_{1}e^{1/t}\left(1-\frac{1}{t}\right)+c_{2}
 $ in $\LaTeX$ 
I am not quite sure how to interpret this answer since there is no function for the $c_2$  does this mean that $c_{1}e^{1/t}\left(1-\frac{1}{t}\right)$ is the only solution, plus or minus a constant?
All of my written work is below, did I make a mistake?
Let $y_{2}=v/t$
 , and we calculate its derivatives (I was using Euler's notation)
$$D_{t}\left[\frac{v}{t}\right]=\frac{v't-v}{t^{2}}$$
and \begin{align*}D_{t}^{2}\left[\frac{v}{t}\right] &= D_{t}\left[\frac{v't-v}{t^{2}}\right]\\
 &= \frac{t^{2}v''-2tv'+2v}{t^{3}}\end{align*}
 Now, we $plug y_{2}$ and its derivatives back into the equation. We get 
\begin{align*}
t^{2}y''+3ty'+y &= 0\\
t^{2}\left[\frac{t^{2}v''-2tv'+2v}{t^{3}}\right]+3t\left[\frac{v't-v}{t^{2}}\right]+v/t &= 0\\
\left[\frac{t^{4}v''-2t^{3}v'+2t^{2}v}{t^{3}}\right]+\left[\frac{3v't^{2}-3vt}{t^{2}}\right]+v/t &= 0\\
tv''-2v'+2\frac{v}{t}+3v'-3\frac{v}{t}+\frac{v}{t} &= 0\\
tv''+-2v'+3v' &= 0\\
tv''+v' &= 0\\
\end{align*}
Again $tv''+v'=0$
  is a first order linear differential equation, which we can solve by separation of parts.
\begin{align*}
tv''+v' &= 0\\
v''+\frac{v'}{t} &= 0\\
\frac{dv'}{v'} &= -\frac{v'}{t}\\
\int\frac{dv'}{v'} &= \int\frac{dt}{t}\\
\ln v' &= \ln t+c\\
v' &= e^{c}t\\
\end{align*}
So, we have that $v'=ct$
 , if we integrate again we have that 
\begin{align*}
\int v'dt &= c\int tdt\\
v &= ct^{2}+k\\
\end{align*}
Lastley we multiply $v$
   by $y_{1}$
 to get 
\begin{align*}
y_{2}(t) &= t^{-1}\left[ct^{2}+k\right]\\
 &= ct+k/t\\
\end{align*}
In the end the last term will be combined with $c_{2}y_{1}$
 , and that the constant $c$
  will get combined with $c_{2}$
  so we have that $y_{2}(t)=t$
   but this cannot be true because y_{2}
  wouldn't be twice differentialable.

For the sake of simplicity it is better to write $y_2=vt^{-1}$ so that $y_2'= v't^{-1}-vt^{-2}$ and $y''_2= v''t^{-1}-2v't^{-2}+2vt^{-3}$. Then 
$t^{2}y_2''+3ty_2'+y_2=tv''+v'=0$. This implies $(tv')'=0$. Integrating this we get $tv'=c_1$. Thus integrating once more we obtain $v=c_1\ln t+c_2$. Since we wish a particular solution which is linearly independent with $y_1=t^{-1}$, we can take $c_1=1$ and $c_2=0$ so that $v=\ln t$ and hence $y_2= \frac{\ln t}{t}$. In your solution you forget the minus sign while you are integrating where $v'$ will be $e^ct^{-1}$.
