Question:

Step 1: Rewrite the differential equation in separated form. Starting with the given differential equation: $$\frac{dy}{dx} = \frac{x^2 + y^2 + 1}{2xy}$$ Separate variables by cross-multiplying: $$2xy\,dy = (x^2 + y^2 + 1)\,dx$$ Step 2: Rearrange to isolate terms involving $y^2$. Rearrange the equation to group terms strategically: $$2xy\,dy - y^2\,dx = (x^2 + 1)\,dx$$ Step 3: Recognize the differential form and make a substitution. Notice that $d(y^2) = 2y\,dy$. Let $v = y^2$, so $dv = 2y\,dy$. Substituting into the rearranged equation: $$x\,dv - v\,dx = (x^2 + 1)\,dx$$ Step 4: Convert to a standard linear differential equation. Divide both sides by $x\,dx$: $$\frac{dv}{dx} - \frac{v}{x} = \frac{x^2 + 1}{x}$$ This can be rewritten as: $$\frac{dv}{dx} - \frac{v}{x} = x + \frac{1}{x}$$ This is a first-order linear ODE in standard form. Step 5: Find and apply the integrating factor. The integrating factor is: $$\mu(x) = e^{-\int \frac{1}{x}\,dx} = e^{-\ln x} = \frac{1}{x}$$ Multiply the entire equation by $\mu(x) = \frac{1}{x}$: $$\frac{d}{dx}\left(\frac{v}{x}\right) = \frac{1}{x}\left(x + \frac{1}{x}\right) = 1 + \frac{1}{x^2}$$ Step 6: Integrate both sides. Integrating both sides with respect to $x$: $$\frac{v}{x} = \int\left(1 + \frac{1}{x^2}\right)\,dx = x - \frac{1}{x} + C$$ Therefore: $$v = x^2 - 1 + Cx$$ Substituting back $v = y^2$: $$y^2 = x^2 - 1 + Cx$$ Step 7: Apply the initial condition to find the constant $C$. Using the initial condition $y(1) = 0$: $$0 = (1)^2 - 1 + C(1)$$ $$0 = 1 - 1 + C$$ $$C = 0$$ Step 8: Write the final solution. Substituting $C = 0$ back into the equation: $$y^2 = x^2 - 1$$ This is the equation of a hyperbola in standard form. **Final Answer:** The solution is $y^2 = x^2 - 1$, which represents a **hyperbola**. The answer is **Option 3: hyperbola**. <div class="key-concept"><strong>Key Concept:</strong> Solving a first-order linear ODE by substitution $v = y^2$ and using an integrating factor</div> <div class="trap-box"><strong>Trap:</strong> Misidentifying the curve type after solving, or making an arithmetic error when applying the initial condition.</div>