Integral curve satisfying y prime=(x2+y2/x2-y2), y(1) =2, has the slope of the curve at the point (1,2), is equal to

Question

Integral curve satisfying y prime=(x2+y2/x2-y2), y(1) =2, has the slope of the curve at the point (1,2), is equal to (A) -5 / 3 (B) -1 (C) 1 (D) 5 / 3

Tardigrade Admin · Accepted Answer

y prime=(x2+y2/x2-y2) y(1,2) prime=(1+4/1-4)=(-5/3)

Q. Integral curve satisfying $y^{'} = \frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}, y (1) = 2$ , has the slope of the curve at the point $(1, 2)$ , is equal to

$- 5/3$

$- 1$

1

$5/3$

$y^{'} = \frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}$
$y_{(1, 2)}^{'} = \frac{1 + 4}{1 - 4} = \frac{- 5}{3}$

Q. Integral curve satisfying y′=x2−y2x2+y2​,y(1)=2, has the slope of the curve at the point (1,2), is equal to

−5/3

−1

1

5/3

y′=x2−y2x2+y2​ y(1,2)′​=1−41+4​=3−5​

Q. Integral curve satisfying $y^{'} = \frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}, y (1) = 2$ , has the slope of the curve at the point $(1, 2)$ , is equal to

$- 5/3$

$- 1$

$5/3$

$y^{'} = \frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}$
$y_{(1, 2)}^{'} = \frac{1 + 4}{1 - 4} = \frac{- 5}{3}$