Solving Riccati equation $y^{\prime}=x^2-y^2$. I was trying to draw the solution curve of the Riccati equation $y^{\prime}=x^2-y^2$, and I met some problem:
On the one hand, I tried substitution
$$
\left\{\begin{array}{l}
u=x+y
\\ v=x-y
\end{array}\right.
,
$$
which leads to
$$
\dfrac{\mathrm du}{\mathrm dv}
=\dfrac{\mathrm du}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}\cdot\dfrac{\mathrm dx}{\mathrm dv}
=\dfrac{\mathrm dy}{\mathrm dx}
=(x+y)(x-y)
=uv. 
$$
This is separable and the solutions are
$$
u=0\text{ and }\ln\left|\dfrac{u}{C}\right|=\dfrac{1}{2}v^2\Rightarrow y=-x\text{ and }x+y=C\mathrm e^{\frac{(x-y)^2}{2}},
$$
where $C$ is an arbitrary constant.

On the other hand, I tried to draw the slope field and the curve in the picture above does not fit the field, since on $y=x$ the slope should be $0$.
I wonder if I've made any mistake. Any help would be appreciated:)

$$y'=x^2-y^2$$
Why not the usual way to solve Riccati ODEs ?
Let $y=\frac{u'}{u}$
$$y'=\frac{u''}{u}-\frac{(u')^2}{u^2}=x^2-\left(\frac{u'}{u}\right)^2$$
$$u''-x^2u=0$$
This is a second order ODE of Bessel kind. See https://mathworld.wolfram.com/BesselDifferentialEquation.html From Eqs.(6-7) the solution is :
$$u=c_1x^{1/2}I_{1/4}(x^2/2)+c_2x^{1/2}I_{-1/4}(x^2/2)$$
$I_{\nu}(z)$ denotes the modified Bessel functions of first kind https://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Then differentiate wrt $x$ to find $u'(x)$ and $y(x)=u'(x)/u(x)$ .
Note :  in case of $I_{\pm 1/4}$ one can use the Parabolic Cylinder functions. https://mathworld.wolfram.com/ParabolicCylinderFunction.html
