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
$\frac{dy}{dx}=\frac{x^2+y^2}{x^2-y^2}⇒\frac{dy}{dx}|_{at\,(1,0)}=1$
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:
Correct answer: option 3
$\frac{dy}{dx}=\frac{x^2+y^2}{x^2-y^2}⇒\frac{dy}{dx}|_{at\,(1,0)}=1$