Prove that the o.d.e. $(\frac{3}{2}\sqrt{|y|}+1+x^2)\frac{dy}{dx}+2xy=0$ has unique local solutions with $y(x_0) = y_0$ for any $x_0$ and $y_0$.
Does the existence and uniqueness theorem for o.d.e's apply?
Attempt:
In my book it states that for an exact equation (an o.d.e. of the form $\frac{dy}{dx} = \frac{-N(x, y)}{M(x,y)}$ where $M = F_y$ and $N = F_x$ for some $C^1$ function $F(x, y)$) the o.d.e. was equivalent to the implicit solution $F(x,y) = c$. So by applying the implicit function theorem, we can assert the local existence of solutions for the implicit equation as long as $M(x, y)$ is not zero. In the specific o.d.e. we have, $M(x,y)$ is never zero so there is always a solution. Is this correct? If so, that's great, but does the existence and uniqueness theorem apply?
You're right, the implicit function theorem works in this case.
For the existence and uniqueness theorem I will now rewrite the equation as $$ \frac{\mathrm{d}y}{\mathrm{d}x} = f(x,y), $$ where $$ f(x,y) = \frac{-2xy}{\frac{3}{2}\sqrt{|y|}+1+x^2}. $$ We need to check if $f$ is locally Lipschitz with respect to $y$. However, $$ \frac{\partial f}{\partial y} = -\frac{2x(\frac{3}{4}\sqrt{|y|}+1+x^2)}{(\frac{3}{2}\sqrt{|y|}+1+x^2)^2} $$ is continuous, hence $f\in C^1$ and in particular $f$ is locally Lipschitz and the uniqueness theorem does apply.