Find a general solution of the reducible second-order differential equation $$yy'' + (y')^2 = yy'$$
So I make my obvious $p$ substitutions:
$$y' = p$$
$$y'' = p(dp/dy)$$
Then:
$$yp(dp/dy) + p^2 = yp$$
Then I know I can reduce this equation to use the integrating factor, so:
$$dp/dy + p/y = 1$$
Where: $$I(x) = e^\int(1/y)dy$$
$$I(x) = y$$
And:
$$y(dp/dy) = 1(y)$$
$$py = (y^2)/2$$
And this is where I get lost. I don't know what the correct next step is, and I end up with convoluted answers.
 A: Here is a simple way to do it. Note that
$${\left( {yy'} \right)^\prime } = yy'' + {\left( {y'} \right)^2}$$
hence your equation becomes
$${\left( {yy'} \right)^\prime } = yy'$$
which results in
$$yy' = C{e^x}$$
but again you can notice that
$${1 \over 2}{\left( {{y^2}} \right)^\prime } = yy'$$
and hence you have
$${\left( {{y^2}} \right)^\prime } = 2C{e^x}$$
Integrating once more leads to
$${y^2} = 2C{e^x} + B$$
A: (To answer the question where to go next - as opposed to the great method outlined by H.R) You were close 
$$
yp = \int y dy = \frac{y^2}{2}+C 
$$
please be careful with notation and doing operations simultaneously on left and right of the equals sign. e.g.
$$
p\mathrm{e}^{\int \frac{1}{y}dy} = \int 1\cdot \mathrm{e}^{\int \frac{1}{y}dy}
$$
not
$$
y\dfrac{dp}{dy} = 1\cdot y
$$
(back to the question)
$$
p\mathrm{e}^{\int \frac{1}{y}dy} = \int 1\cdot \mathrm{e}^{\int \frac{1}{y}dy} = \int y dy = \frac{y^2}{2}+C
$$
(this is probably where you forgot the constant along other things)
$$
\frac{yy'}{y^2+2C} = \frac{1}{2}
$$
let $y^2+2C = v$ we find
$$
\frac{v'}{2v} = \frac{1}{2}\implies \ln v = x + C_1
$$
so we have
$$
v = A\mathrm{e}^x = y^2+2C
$$
or
$$
y^2 = A\mathrm{e}^x - 2C
$$
