Question:

<p><strong>Step 1:</strong> Verify homogeneity. Rewrite as: $(x^2 - y^2)dx + 2xy\,dy = 0$. Both terms are degree 2, confirming this is homogeneous.</p><p><strong>Step 2:</strong> Use substitution $y = vx$, so $dy = v\,dx + x\,dv$. Substitute:</p><p>$(x^2 - v^2x^2)dx + 2x(vx)(v\,dx + x\,dv) = 0$</p><p>$x^2(1 - v^2)dx + 2v x^2(v\,dx + x\,dv) = 0$</p><p><strong>Step 3:</strong> Simplify: $(1 - v^2 + 2v^2)dx + 2vx\,dv = 0$</p><p>$(1 + v^2)dx + 2vx\,dv = 0$</p><p><strong>Step 4:</strong> Separate variables: $\frac{dx}{x} + \frac{2v\,dv}{1+v^2} = 0$</p><p><strong>Step 5:</strong> Integrate: $\ln|x| + \ln(1+v^2) = \ln|C|$</p><p>$x(1 + v^2) = C$</p><p><strong>Step 6:</strong> Substitute back $v = \frac{y}{x}$:</p><p>$x\left(1 + \frac{y^2}{x^2}\right) = C$</p><p>$x + \frac{y^2}{x} = C$ or $x^2 + y^2 = Cx$</p><p><strong>Step 7:</strong> Apply initial condition $(1,1)$: $1 + 1 = C(1)$, so $C = 2$</p><p>∴ Answer: $x^2 + y^2 = 2x$ or equivalently $(x-1)^2 + y^2 = 1$ (circle with center (1,0) and radius 1)</p>