Two differential equations - separation of variables, Gronwall's lemma 1) How to solve $xyy' = \ln x$ without separation of variables? That's what I am asked to do.
$$yy' = \frac{\ln x}{x} \Rightarrow \frac{y^2}{2} =\frac{\ln^2 x}{2} + C \Rightarrow \dots$$
Unfortunately this method is not allowed.
2) How can one using the Gronwall's lemma prove that $y(t) = -1$ is the only solution of $y' = t(y+1)$, $y(0) = -1$?
 A: Another way for the first problem:
Try to use the change of variables. Put $z = \ln x$. Then $x = e^{z}$ and
for $y(z) = y(x(z))$ you will get
$$
y'_z = y'_x x'_z = y'_x e^z \quad \Longrightarrow \quad y'_x = \frac{y'_z}{e^z}.
$$
Substituting this into your differential equation, you will get
$$
y y'_z = z.
$$
Then proceed as usual, without separation of variables.
A: A solution to the second problem: 
Let $T>0$ $(T$ is arbitrary$)$. We will show that the solution is unique for $t\in[0,T]$. Let's suppose that there is another solution: $y_1$. Then:
$$|y_1(t)-y(t)|=\left|\int_{0}^{t}s(1+y_1(s))-s(1+y(s))ds\right|=\left|\int_{0}^{t}s(y_1(s)-y(s))ds\right|\leq\int_{0}^{t}|s||y_1(s)-y(s)|ds\leq T\int_{0}^{t}|y_1(s)-y(s)|ds$$
And now let $u=|y_1(t)-y(t)|$. Then $u$ satisfies the conditions of Gronwall's lemma with $a=0$. So $u=0$. So the solution is unique.
A: Note that $x>0$
  for your first equation. Then, $y\left(x\right)=\ln\left(x\right)$
  is such that $$y'\left(x\right)y\left(x\right)=\frac{\ln\left(x\right)}{x}$$
 so $y$
  is a solution of your equation (I do not see what was the problem).
For your second equation, $y\equiv-1$
  is the unique solution by the Cauchy-Lipschitz theorem, and it is a global solution (because constant), and Gronwall lemma is not used here.
