Problem with an integration involving natural log I have a question regarding the following integration:
$\frac{d}{dy}(y\frac{dp}{dy})=-\frac{y}{a}\frac{dp}{dy}$, where $a$ is just a constant. 
My book then tells me that this expression can be solved for $(y\frac{dp}{dy})$, and then get the following result:
$ln(y\frac{dp}{dy})=-\frac{y}{a}+C$, where $C$ is also a constant.
I do not understand how they get from the first expression to the other expression, so I hope someone can help me with that.
David
 A: HINT:
$$\frac{\text{d}}{\text{d}y}\left(y\cdot\frac{\text{d}p}{\text{d}y}\right)=-\frac{y}{a}\cdot\frac{\text{d}p}{\text{d}y}\Longleftrightarrow yp''(y)+p'(y)=-\frac{yp'(y)}{a}$$
Let $p'(y)=r(y)$, so $p''(y)=r'(y)$:
$$r(y)+yr'(y)=-\frac{yv(y)}{a}\Longleftrightarrow r'(y)=-\frac{r(y)(a+y)}{ay}\Longleftrightarrow\frac{ar'(y)}{r(y)}=-\frac{a+y}{y}$$
Now, Integrate both sides with respect to $y$:
$$\int\frac{ar'(y)}{r(y)}\space\text{d}y=\int-\frac{a+y}{y}\space\text{d}y$$
For the LHS, substitute $u=r(y)$ and $\text{d}u=r'(y)\space\text{d}y$:
$$\int\frac{ar'(y)}{r(y)}\space\text{d}y=a\int\frac{1}{u}\space\text{d}u=a\ln|u|+\text{C}=a\ln|r(y)|+\text{C}$$
For the RHS:
$$\int-\frac{a+y}{y}\space\text{d}y=-\left[a\int\frac{1}{y}\space\text{d}y+\int1\space\text{d}y\right]=-\left(a\ln|y|+y\right)+\text{C}$$
So, we get:
$$a\ln|p'(y)|=-\left(a\ln|y|+y\right)+\text{C}$$
A: Let $z = y\frac{dp}{dy}$.  Then your equation is $\frac{dz}{dy} = -\frac{z}{a}$ which is separable:  $\frac{dz}{z} = -\frac{dy}{a}$.  Integrate both sides to get $\ln z = -\frac{y}{a} + C$.  Plug back in for $z$.
