Let the solution curve of the differential equation $$xdy-ydx=\sqrt{x^{2}+y^{2}}dx,x>0$$, $$y(1) = 0$$ , be $$y$$ = $$y(x)$$ . Then $$y(3)$$ is equal to

$$x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx \implies \frac{dy}{dx} - \frac{y}{x} = \sqrt{1 + \left(\frac{y}{x}\right)^2}$$

$$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$

$$v + x\frac{dv}{dx} - v = \sqrt{1 + v^2} \implies x\frac{dv}{dx} = \sqrt{1 + v^2}$$

$$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$$

$$\ln\left\vert{}v + \sqrt{1+v^2}\right\vert{} = \ln\vert{}x\vert{} + \ln C \implies \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = Cx$$

$$0 + \sqrt{1 + 0} = C(1) \implies C = 1$$

Solution curve: $$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = x \implies y + \sqrt{x^2 + y^2} = x^2$$

$$y(3) + \sqrt{9 + [y(3)]^2} = 9$$

$$\sqrt{9 + [y(3)]^2} = 9 - y(3)$$

$$9 + [y(3)]^2 = 81 - 18y(3) + [y(3)]^2$$

$$18y(3) = 72 \implies y(3) = 4$$