Solve the differential equation $\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$, if $y(1)=0$ I don’t know what form this. It’s probably a reducible to linear form equation, but I don’t know which term to substitute or how to rearrange. Can I get a hint?
 A: Hint; Put $x\log x=t$
we have $$\frac{dt}{dy}-t=e^y$$
$$e^{-y}dt-te^{-y}dy=1$$
$$d(e^{-y}t)=1$$
A: We need to solve
\begin{eqnarray*}
(IVP): \left\{\begin{aligned} \frac{dx}{dy}-\frac{x\ln(x)}{1+\ln(x)}=\frac{e^{y}}{1+\ln(x)}\\
y(1)=0 \\ \end{aligned}\right.\\
\end{eqnarray*}
Now, rewrite the ODE, you have $$e^{y}+x\ln(x)-(\ln(x)+1)\frac{dx}{dy}=0$$
(If you have difficulty getting to this part, tell me to put the details).
Let $P(y,x)=e^{y}+x\ln(x)$ and $Q(y,x)=-\ln(x)-1$, so we can see this ODE is not an exact equation, because $P_{x}=\ln(x)+1\not=0=Q_{y}$.
Now, find an integrating factor $\mu(y)$ such that $$\mu(y)R+\frac{dx}{dy}\mu(y)Q=0$$ is exact.
Solving this part, we can see that $\mu(y)=e^{-y}$, and multiply both sides of $$e^{y}+x\ln(x)+(-1)(1+\ln(x))\frac{dx}{dy}=0$$ by $\mu(y)$ we have a exact equation.
Now, using the method for the method to solve exact equations (if you don't know this method, please tell me to fill in the details), we have that exists a function $f$ such that $$f_{y}=P \quad f_{x}=Q$$ and $$f(x,y)=c$$
Solving this, we have $$f(y,x)=y-x\ln(x)e^{-y}$$
so, the solution for the ODE is $$\boxed{y-x\ln(x)e^{-y}=c}$$
Now, just use the initial condition $y(1)=0$ so, you have the solution for IVP. Solving this last part we have $c=0$ and the solution for IVP is $$\boxed{y-x\ln(x)e^{-y}=0}$$
A: $$\frac{dx}{dy} -\frac{x \log x}{1+\log x} =\frac{e^y}{1+\log x}$$
$$({1+\log x}){dx} -(x \log x)dy=e^ydy$$
Note that:
$$d(x \log x)=(1+\log x) dx$$
So that we have:
$$d(x \log x) -x \log xdy={e^y}dy$$
Multiply by $e^{-y}$:
$$e^{-y}d(x \log x)+ x \log xde^{-y}=dy$$
$$d(e^{-y}x \log x)=dy$$
Integrate:
$$e^{-y}x \log x-y=C$$
