Why is $y$ squared?

$\begin{aligned} \frac{dy}{dx} & = \frac{\sqrt{x^2+y^2}-x}{y} \\ y \frac{dy}{dx} + x & = \sqrt{x^2+y^2} \\ \frac{d}{dx} \left(\frac{y^2}{2} + \frac{x^2}{2} \right) & = \sqrt{x^2+y^2} \\ \frac{d}{dx} \left(\color{#3D99F6}{y^2 + x^2} \right) & = 2 \sqrt{\color{#3D99F6}{x^2+y^2}} \quad \quad \small \color{#3D99F6}{\text{Let } z = x^2+y^2} \\ \frac{d\color{#3D99F6}{z}}{dx} & = 2 \sqrt{\color{#3D99F6}{z}} \\ \int z^{-\frac{1}{2}} \space dz & = \int 2 \space dx \\ 2 z^{\frac{1}{2}} & = 2 x + 2\color{#3D99F6}{c} \quad \quad \small \color{#3D99F6}{c = \text{a constant}} \\ z^{\frac{1}{2}} & = x + c \\ \Rightarrow \sqrt{x^2+y^2} & = x + c \quad \quad \small \color{#3D99F6}{\text{When }x = 0, y = 4} \\ \sqrt{0^2+4^2} & = 0 + c \\ \Rightarrow c & = 4 \\ \Rightarrow \sqrt{x^2+y^2} & = x + 4 \\ x^2+y^2 & = x^2 + 8x + 16 \\ \Rightarrow y^2 & = 8x + 16 \\ \text{For } y = 0: \quad 8(\beta+2) & = 0 \\ \Rightarrow \beta & = \boxed{-2} \end{aligned}$

Subsituting $y = xu$ , the DE becomes $xu' + u \; =\; \frac{\sqrt{1+u^2}-1}{u} \qquad \mbox{or} \qquad xu' \; = \; \frac{\sqrt{1+u^2}(1-\sqrt{1+u^2})}{u} \;,$ and so, separating variables, $0 \; = \; \int \frac{dx}{x} + \int \frac{u\,du}{\sqrt{1+u^2}(\sqrt{1+u^2}-1)} \;.$ Putting $v = \sqrt{1+u^2}$ yields $0 \; = \; \int \frac{dx}{x} + \int \frac{dv}{v-1} \; = \; ln x + \ln (v-1) + c$ (choosing $v-1$ rather than $|v-1|$ , since this will match the initial condition shortly to be applied). Thus $A \; = \; x(v-1) \; = \; x\sqrt{1+u^2} - x \; = \; \sqrt{x^2 + y^2} - x \;.$ Since $y(0) = 4$ we deduce that $A=4$ , and so $\sqrt{x^2 + y^2} = x + 4$ .

Putting $y=0$ , we need to solve $|x| = x+4$ . The only solution to this equation is $x=-2$ .

Make the substitution $r^2=x^2+y^2$

By implicit differentiation, this gives

$\frac{d}{dx} r^2=\frac{d}{dx}( x^2+y^2)$ $\Rightarrow 2r\frac{dr}{dx}=2x+2y\frac{dy}{dx}$ $\Rightarrow \frac{dy}{dx}=\frac{r\frac{dr}{dx}-x}{y}$

Hence, the original equation reduces to

$\frac{r\frac{dr}{dx}-x}{y}=\frac{\sqrt{r^2}-x}{y}$ (No need to consider absolute value of $r$ since $x^2+y^2$ is always possitive) $\Rightarrow r\frac{dr}{dx}=r\Rightarrow \frac{dr}{dx}=1$ This is a separable ODE, which yields $\int dr=\int dx$ $\Rightarrow r=x+C \Rightarrow x+C=\sqrt{x^2+y^2}$ Hence, the general solution is $y^2-2Cx-C^2=0$ Since $y(0)=4$ , we know that $C=\pm4$ , but for there to be one NEGATIVE solution to $y(x)=0$ , we choose $C=4$ . Hence: $y^2=8x+16$ Finally, we solve $y(x)=0\Rightarrow 8x+16=0$ $\boxed{x=-2}$