I am at my first attempts to do qualitative analysis of a differential equation and I found myself stuck at this exercise:
Study the solutions of $y'=\frac{x}{1+\log(y)}$
These are my attempts:
- I found the isocline curves which are given by $y(x_0)=e^{\frac{x_o}{c}-1}$;
- I found that the second derivative of $y$, namely $y''=\frac{1}{1+\log(y)}-\frac{x^2}{y(1+\log(y))^3}$;
But I'm stuck here because studying the sign of $y''$ isn't easy. Can you suggest me how to proceed further? Any suggestions about general cases (hints, things to check before starting the study, standard method for studying determined cases) is really welcome. Thanks in advance.
This differential equation can be solved as
$$ (1+\ln(y))dy = x dx $$
or after integration
$y\ln y = \frac{1}{2}x^2+ C$
Now calling $y = e^z$ we have
$$ z e^z = \frac{1}{2}x^2+ C $$
then using the Lambert function $W(\cdot)$ we have
$$ z = W\left(\frac{1}{2}x^2+ C\right) = \ln y $$
or
$$ \displaystyle y = e^{W\left(\frac{1}{2}x^2+ C\right)} $$
Attached a plot showing the phase plane and a integral for $C = 0$