Integral curve satisfying $y'=\frac{x^2+y^2}{x^2-y^2}, y(1)=1$ has the slope at the point (1, 0) of the curve, equal to :
Answer & explanation
Correct answer: option 3
$\frac{d y}{d x}=\frac{x^2+y^2}{x^2-y^2}$
⇒ slope at (1,0) = $\left[\frac{d y}{d x}\right]_{(1,0)}=\frac{1+0}{1-0}=1$
Hence (3) is the correct answer.