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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 |
$-\frac{5}{3}$ -1 1 $\frac{5}{3}$ |
1 |
We have, $\frac{d y}{d x}=\frac{x^2+y^2}{x^2-y^2}$ ∴ Slope of the tangent at $(1,0)=\left(\frac{d y}{d x}\right)_{(1,0)}=\frac{1+0}{1-0}=1$ |
