The differential equation above is a Ricatti equation. I'm not given any particular solution to it. I'm actually asked to solve it. I tried graphing the direction field to get an idea on the behavior of the curve with no result.
However, it is true that this equation has no real solution. To solve it we need to use Bessel equations (I'm still being introduced to second order differential equations).
How can we prove that $2xy'=4x^{2}+3y^{2}$ has no real solution? Can we use the existence and unicity theorem?
Also, it is possible to prove it without using Lipschitz continuity?
The equation
$2xy' = 4x^2 + 3y^2 \tag 1$
has a local real solution through any $(x_0, y_0)$, where $y_0 = y(x_0)$, provided $x_0 \ne 0$; to see this, note that about any $x_0 \ne 0$ there is an open interval $I = (x_0 - \delta, x_0 + \delta)$, where $\delta < \vert x_0 \vert$, such that
$x \in I \Longrightarrow x \ne 0; \tag 2$
then on $I$ we may write (1) as
$y' = 2x + \dfrac{3y^2}{2x} = f(x, y); \tag 3$
since $f(x, y)$ is differentiable for $(x, y) \in I \times \Bbb R$, it is both continuous in $x$ and locally Lipschitz continuous in $y$ on $I \times \Bbb R$; this fact is quite well-known and has been discussed more than once on this site. Since $f(x, y)$ is continuous in $x$ and locally Lipschitz continuous in $y$ on $I \times \Bbb R$, the Picard-Lindeloef theorem shows that (2) has a unique real solution through any $(x_0, y_0) \in I \times \Bbb R$; hence does (1) since $x \ne 0$ on $I$.
I don't know how to show (1)-(2) has unique solutions where $x \ne 0$ without Picard-Lindeloef, which requires $f(x, y)$ to meet the criterion given in the previous paragraph.
On the other hand, and I haven't the time to analyze this in detail at the moment, there may be challenges in trying to extend a solution to/through the $x$-value $x = 0$. So perhaps that is what is meant by saying there are no real solutions--there may not be on all of $\Bbb R$.
Perhaps someone else can chime in here . . .