CUET Preparation Today
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 : |
$-\frac{5}{3}$ -1 1 $\frac{5}{3}$ |
1 |
$\frac{d y}{d x}=\frac{x^2+y^2}{x^2-y^2}$ ⇒ 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. |
