Question:
If a cuver passes through the point $(1,-2)$ and has slope of the tangent at any point $(\mathrm{x}, \mathrm{y})$ on it as
$\frac{x^{2}-2 y}{x}$, then the curve also passes through the point :
Correct Option: , 2
Solution:
$\frac{d y}{d x}=\frac{x^{2}-2 y}{x}$ (Given)
$\frac{d y}{d x}+2 \frac{y}{x}=x$
$\mathrm{I} F=e^{\int \frac{2}{x} d x}=x^{2}$
$\therefore y \cdot x^{2}=\int x \cdot x^{2} d x+C$
$=\frac{x^{4}}{y}+C$
hence bpasses through $(1,-2) \Rightarrow \mathrm{C}=-\frac{9}{4}$
Now check option(s), Which is satisly by option (ii)