Differential Equations
Population ODE
Grade Class 12
Question:
<p>\\((x^2-y^2)\\,dx+2xy\\,dy=0\\) through \\((1,1)\\). Find the curve.</p>
<span>\(x^2+y^2 = 2x\)</span>
<span>\(x^2-y^2 = x\)</span>
<span>\(x^2+y^2 = 3x\)</span>
<span>\(2(x^2+y^2) = 3x\)</span>
Step-by-Step Solution
Key Concept: Homogeneous or exact ODE — try grouping.
<div class='solution'><p><strong>Step 1:</strong> Rearrange: $x^2\,dx - y^2\,dx + 2xy\,dy = 0$ → $x^2\,dx + d(xy^2) - y^2\,dx = 0$... Try grouping: $(x^2-y^2)dx+2xy\,dy=0$. This is homogeneous (degree 2).</p><p><strong>Step 2:</strong> Let $y=vx$: $(1-v^2)+2v(v+xv')=0$ → $1+v^2+2xvv'=0$ → $2xvv'=-(1+v^2)$ → $\dfrac{2v\,dv}{1+v^2}=-\dfrac{dx}{x}$.</p><p><strong>Step 3:</strong> $\ln(1+v^2)=-\ln x+C$ → $x(1+v^2)=k$ → $x+y^2/x=k$ → $x^2+y^2=kx$. Through $(1,1)$: $k=2$. <strong>Answer: (1)</strong> $x^2+y^2=2x$.</p></div>
Correct Answer: 1