General solution of the differential equation \( x(xdy-ydx)=y\,dx \) is: (where c is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \left|\displaystyle\frac{y}{x}\right|=e^{\frac{-1}{x}+c} \)
Expanding and collecting the dx terms, the variables separate.
\( \displaystyle x^2\,dy-xy\,dx=y\,dx \)
\( \displaystyle x^2\,dy=y(x+1)\,dx \)
\( \displaystyle \frac{dy}{y}=\frac{x+1}{x^2}\,dx \)
\( \displaystyle =\left(\frac{1}{x}+\frac{1}{x^2}\right)dx \)
Integrating both sides, with \( \int\frac{dx}{x^2}=-\frac{1}{x} \):
\( \displaystyle \log|y|=\log|x|-\frac{1}{x}+c \)
\( \displaystyle \log\left|\frac{y}{x}\right|=-\frac{1}{x}+c \)
\( \displaystyle \left|\frac{y}{x}\right|=e^{\frac{-1}{x}+c} \)
Why the other options are incorrect
Option (1) \( \left|\displaystyle\frac{1}{x}\right|=e^{\frac{-y}{x}+c} \) — Incorrect. Its left side does not contain y, and the exponent \( \frac{-y}{x} \) does not come from the integration; the separated equation gives \( \log\left|\frac{y}{x}\right| \) on the left.
Option (2) \( \left|\displaystyle\frac{1}{x}\right|=e^{\frac{y}{x}+c} \) — Incorrect. Its left side does not contain y either; the separated equation gives \( \log\left|\frac{y}{x}\right| \) on the left.
Option (4) \( \left|\displaystyle\frac{y}{x}\right|=e^{\frac{1}{x}+c} \) — Incorrect. This uses \( \int\frac{dx}{x^2}=\frac{1}{x} \); the correct integral is \( -\frac{1}{x} \).