The general solution of the differential equation \((x^2 + y^2)\,dy = 2xy\,dx\), \(x \ne 0\), \(y \ne 0\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x^2 - y^2 = Cy\)
The equation is
\( \displaystyle \frac{dy}{dx} = \frac{2xy}{x^2 + y^2} \)
a homogeneous equation. Let \(y = vx\):
\( \displaystyle v + x\frac{dv}{dx} = \frac{2v}{1 + v^2} \)
\( \displaystyle x\frac{dv}{dx} = \frac{2v - v - v^3}{1 + v^2} = \frac{v(1 - v^2)}{1 + v^2} \)
\( \displaystyle \frac{1 + v^2}{v(1 - v)(1 + v)}\,dv = \frac{dx}{x} \)
By partial fractions:
\( \displaystyle \frac{1 + v^2}{v(1 - v)(1 + v)} = \frac{1}{v} + \frac{1}{1 - v} - \frac{1}{1 + v} \)
Integrating:
\( \log|v| - \log|1 - v| - \log|1 + v| = \log|x| + \log|k| \)
\( \displaystyle \frac{v}{1 - v^2} = kx \)
Putting \(v = \frac{y}{x}\): \(\frac{v}{1 - v^2} = \frac{xy}{x^2 - y^2}\), so
\( \displaystyle \frac{xy}{x^2 - y^2} = kx \)
\( \displaystyle x^2 - y^2 = \frac{1}{k}\,y = Cy \)
Why the other options are incorrect
Option (1) \(x^2 + y^2 = Cy\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = \frac{2xy}{x^2 - y^2}\), with a minus sign in the denominator.
Option (3) \(x^2 - y^2 = Cx\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\), the reciprocal of the given slope.
Option (4) \(x^2 + y^2 = Cx\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\), not \(\frac{2xy}{x^2 + y^2}\).