<s>[INST] Solve the initial value problem $xyy' + xy' = 1$ and $y(1) = 0$ I can solve the differential equation, which is $y + y^{2}/2 = \ln(x) + C$.
But I cannot solve the IVP because I can't isolate for $y$ and find the value of $C.$
 [/INST] We have (just reproducing your work for completeness, and combining some of the comment ideas):
\begin{align*}
xyy'+xy'&=1\\
xy'(y+1)&=1\\
(y+1)\,dy&=\frac{dx}{x}\\
\frac{y^2}{2}+y&=\ln(x)+C\\
0&=\ln(1)+C\\
0&=C\\
\frac{y^2}{2}+y&=\ln(x)\\
y^2+2y&=2\ln(x)\\
y^2+2y+1&=2\ln(x)+1\\
(y+1)^2&=2\ln(x)+1\\
y+1&=\pm\sqrt{2\ln(x)+1}\\
y&=-1\pm\sqrt{2\ln(x)+1}.
\end{align*}
But notice that we can't allow the negative square root, because it doesn't actually satisfy the initial condition. Also notice that we threw out the $\ln|x|$ for $\ln(x),$ because we knew the initial condition would be for positive $x.$ So the final solution is
$$y=-1+\sqrt{2\ln(x)+1}. $$
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