Integral curve satisfying $\frac{d y}{d x}=\frac{x^2+y^2}{x^2-y^2}, y(1)=2$ has the slope at the point $(1,0)$ of the curve equal to
Answer & explanation
Correct answer: option 3
We have, $\frac{d y}{d x}=\frac{x^2+y^2}{x^2-y^2}$
∴ Slope of the tangent at $(1,0)=\left(\frac{d y}{d x}\right)_{(1,0)}=\frac{1+0}{1-0}=1$