<s>[INST] Solve $y''(1+\ln(x)) + \frac{1}{x}y' = 2+\ln(x)$ When $y = \frac{1}{2}$, $y' = 1$, for $x=1$ Solve $$y''(1+\ln(x)) + \frac{1}{x}y' = 2+\ln(x)$$
First I converted it to $$p' +\frac{1}{x(1+\ln(x))}y' = \frac{2+\ln(x)}{1+\ln(x)} $$
Which looks like a nice first order differential equation. However, when $I = 1+ \ln(x)$ you solve for $p$ you get the following DE:
$$y' = \frac{2x+\frac{1}{x}+C}{1+\ln(x)}$$
Which does not produce the right result when I put it through a calculator. Where is the mistake?
 [/INST] I think your error was the antiderivative of $\ln x$: it is $x\ln x - x$. Here is a complete computation of the result:
Note that
$$
y''(1+\ln x) + \frac{1}{x} y' = \left[y' (1 + \ln x) \right]',
$$
and the equation becomes
$$
\left[y'(1+\ln x)\right]' = 2 + \ln x.
$$
Integrating, it follows that
$$
y' (1+\ln x) = 2x + x\ln x - x + C = x (1 + \ln x) + C,
$$
and
$$
y' = x + \frac{C}{1 + \ln x}.
$$
If $y'(1) = 1$, then $C = 0$ and $y' = x$. Finally, $y(x) = \frac{x^2}{2} + C'$, and $y(1) = \frac{1}{2}$ yields
$$
y(x) = \frac{x^2}{2}.
$$
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