Using reduction of order to find general solution of DE I have been given the following DE...
$$ ty' ' -  ( 1 + 3  t ) y '    +  3   y = 0 \ ,   \ \ t \neq 0 $$
The problem states that the DE has a solution of form $ e^{ct} $, where $c$ is some constant. Now, we need to find the general solution.
I used reduction of order like so...I first set $y_2 = v(t)y_1$, where $v$ is some function in terms of t. First, I modified the DE by dividing t from both sides, like so:
$$ y''-\frac{1+3t}{t}y'+\frac{3}{t}y=0 $$
Alright, now:
$$ p(t)=-\frac{1+3t}{t} $$
I use a formula for reduction of order, which is:
$$ v(t)=\int u(t) \ dt $$
where $u(t)$ is equal to:
$$ u(t)=\frac{e^{-\int p(t) \ dt}}{(y_1)^2} $$
Therefore, plugging in values, I get, where $c_1$ and $c_2$ are constants:
$$ y=c_1e^{ct}+c_2(\int te^{3t-2ct}dt)e^{ct}$$
Of course, to come to my problem. I cannot really integrate until I find what $c$ is. How can I find $c$? Thank you!
 A: As commented by  jdods earlier, let $$y= e^{ct}\,z \qquad y'=e^{c t} \left(c z+z'\right)\qquad y''=e^{c t} \left(c^2 z+2 c z'+z''\right)$$ and replace to get
$$e^{c t} \left(t z''+(2 c t-3 t-1) z'+(c-3) (c t-1) z\right)=0$$ that is to say
$$t z''+(2 c t-3 t-1) z'+(c-3) (c t-1) z=0$$
 To reduce simply the order, let the coefficient of $z$ to be $0$.
A: Let $k$ be an arbitrary constant.  We shall solve the second-order differential equation of the form
$$t\,y''(t)-(kt+1)\,y'(t)+k\,y(t)=0\,.$$
Observe that the equation can be written in the form
$$t\,\big(y''(t)-k\,y'(t)\big)-\big(y'(t)-k\,y(t)\big)=0\,.$$
Therefore,
$$\frac{\text{d}}{\text{d}t}\,\left(\frac{y'(t)-k\,y(t)}{t}\right)=0\text{ or }y'(t)-k\,y(t)=a\,t\,,$$
for some constant $a$.  Hence,
$$\frac{\text{d}}{\text{d}t}\,\big(\exp(-k\,t)\,y(t)\big)=a\,t\,\exp(-k\,t)\,.$$
If $k=0$, then 
$$y(t)=\frac{a}{2}\,t^2+b\text{ for some constant }b\,.$$
If $k\neq 0$, then there exists a constant $b$ such that
$$y(t)=\exp(k\,t)\,\left(-a\,\frac{(k\,t+1)\,\exp(-k\,t)}{k^2}+b\right)=-\frac{a}{k^2}\,(kt+1)+b\,\exp(k\,t)\,.$$
