allerySolve $(x^2 - y^2)\,dx + 2xy\,dy = 0$.
A$x^2 + y^2 = cx$ (family of circles through origin)
B$x^2 - y^2 = cx$
C$xy = c$
D$x + y = c$
Answer & Solution
Correct answer: A. $x^2 + y^2 = cx$ (family of circles through origin)
Homogeneous.
Substitute $y = vx$: $(1 - v^2)\,dx + 2v(v\,dx + x\,dv) = 0$
⇒ $(1 - v^2 + 2v^2)\,dx + 2vx\,dv = 0$
⇒ $(1 + v^2)\,dx + 2vx\,dv = 0$
⇒ $dx/x + 2v\,dv/(1+v^2) = 0$.
**Integrating:** $\log x + \log(1+v^2) = \log c$
⇒ $x(1 + v^2) = c$ ⇒ $x + y^2/x = c$ ⇒ $x^2 + y^2 = cx$.
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