I am really struggling with this problem. This section of my book is about solutions of differential equations through substitutions. We were taught about three methods of substitution in this section.
The first about Homogeneous equations. where you have something like
$$ M(x,y) + N(x,y) = 0$$
and
$$M(tx,ty) = t^\alpha *M(x,y)\; and\; N(tx,ty) = t^\alpha *N(x,y)$$ then using the substitution u=y/x or v=x/y
the other method we learned was Bernoullis equation where you have the DE in the form
$$\frac{dy}{dx}+P(x)y = f(x)y^n$$
and the last we learned of was Reduction to separation of variables. Where you have $$\frac{dy}{dx}=f(Ax+By+C)$$ and you use subsitutions
The problem states determine an appropriate substitution and solve
$$xy'=yln(xy)$$
I found an answer on yahoo answers but I am lost in the steps.
https://answers.yahoo.com/question/index?qid=20150420091352AA9lvDl
The answer given is the same as the correct answer. Can anyone explain the steps especially the part where it transitions from $$xydu=xdy+ydx = y(u+1)dx$$ or tell me another approach.Thank you
Edit:
My teacher has told me that this problem was not of three types I mentioned above, which was confusing me as I was set on it having to be one of them. Thank you everyone for your time and solutions.
Best Answer
Hint
$$xy'=y\ln(xy)$$ $$xy'+y=y\ln(xy)+y$$ $$xy'+y=y(\ln(xy)+1)$$ $$(xy)'=\frac {xy(\ln(xy)+1)} x$$ Integrate both sides $$\int \frac {d(xy)} {xy(\ln(xy)+1)}=\int \frac{dx}x$$ Substitute $z=xy$ for simplicity $$\int \frac {dz} {z(\ln(z)+1)}=\int \frac{dx}x$$ Substitute $z=e^t$ and ofcourse $dz=e^tdt$ $$\int \frac {\color{red}{e^t}dt} {\color{red}{e^t}(\ln(e^t)+1)}=\int \frac{dx}x$$ Note that $\ln(e^t)=t$ $$\int \frac {dt} {(t+1)}=\ln(x) +K$$ Don't forget to substitute back $t=\ln(z)=\ln(xy)$ $$\ln(t+1)=\ln(x)+K \implies t+1=Kx \implies \ln(xy)=Kx-1$$ And finally $$\boxed {y(x)=\frac {e^{Kx-1}} x}$$