The general solution of differential equation \( x(x\,dy-y\,dx)=y\,dx \) is (where \( C \) is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \log|y|-\log|x|=-\displaystyle\frac{1}{x}+C \)
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 \( \int\frac{dx}{x^2}=-\frac{1}{x} \):
\( \displaystyle \log|y|=\log|x|-\frac{1}{x}+C \)
\( \displaystyle \log|y|-\log|x|=-\frac{1}{x}+C \)
Why the other options are incorrect
Option (2) \( \log|y|+\log|x|=-\displaystyle\frac{1}{x}+C \) — Incorrect. Moving \( \log|x| \) to the left side changes its sign, so the left side is \( \log|y|-\log|x| \), not \( \log|y|+\log|x| \).
Option (3) \( \log|y|-\log|x|=\displaystyle\frac{1}{x}+C \) — Incorrect. This uses \( \int\frac{dx}{x^2}=\frac{1}{x} \); the correct integral is \( -\frac{1}{x} \).
Option (4) \( \log|y|+\log|x|=\displaystyle\frac{1}{x}+C \) — Incorrect. Both the sign of \( \log|x| \) and the sign of \( \frac{1}{x} \) are wrong.