1. **State the problem:** Solve the differential equation $$x(1 + y^2)^{1/2} \, dx = y(1 + x^2)^{1/2} \, dy$$ and find the explicit solution. 2. **Rewrite the equation:** We can write it as $$x \sqrt{1 + y^2} \, dx = y \sqrt{1 + x^2} \, dy$$. 3. **Separate variables:** Rearranging terms to separate variables, divide both sides by $$\sqrt{1 + y^2} \sqrt{1 + x^2}$$: $$\frac{x}{\sqrt{1 + x^2}} \, dx = \frac{y}{\sqrt{1 + y^2}} \, dy$$. 4. **Integrate both sides:** - Left side: $$\int \frac{x}{\sqrt{1 + x^2}} \, dx$$ - Right side: $$\int \frac{y}{\sqrt{1 + y^2}} \, dy$$ 5. **Evaluate the integrals:** Use substitution for the left integral: let $$u = 1 + x^2$$, then $$du = 2x \, dx$$, so $$x \, dx = \frac{du}{2}$$. Left integral becomes: $$\int \frac{x}{\sqrt{1 + x^2}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot 2 u^{1/2} + C = \sqrt{1 + x^2} + C$$. Similarly for the right integral, let $$v = 1 + y^2$$, then $$dv = 2y \, dy$$, so $$y \, dy = \frac{dv}{2}$$. Right integral becomes: $$\int \frac{y}{\sqrt{1 + y^2}} \, dy = \int \frac{1}{\sqrt{v}} \cdot \frac{dv}{2} = \frac{1}{2} \int v^{-1/2} \, dv = \sqrt{1 + y^2} + C$$. 6. **Set the integrals equal:** $$\sqrt{1 + x^2} = \sqrt{1 + y^2} + C$$. 7. **Rewrite the implicit solution:** $$\sqrt{1 + x^2} - \sqrt{1 + y^2} = C$$. This is the implicit general solution. 8. **If initial conditions are given, solve for C:** Since no initial condition is provided, this is the final implicit solution. **Final answer:** $$\sqrt{1 + x^2} - \sqrt{1 + y^2} = C$$