Differential Equations Question 114
Question: The slope of the tangent at (x, y) to a curve passing through a point (2, 1) is $ \frac{x^{2}+y^{2}}{2xy} $ , then the equation of the curve is
[MP PET 2002]
Options:
A) $ 2(x^{2}-y^{2})=3x $
B) $ 2(x^{2}-y^{2})=6y $
C) $ x(x^{2}-y^{2})=6 $
D) $ x(x^{2}+y^{2})=10 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy} $ . Put $ y=vx $
Therefore $ v+x.\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{2vx^{2}} $
$ \frac{2v}{1-v^{2}}.dv=\frac{dx}{x} $
Integrating both sides, $ -\log (1-v^{2})=\log x+\log c $
$ -\log ( 1-\frac{y^{2}}{x^{2}} )=\log x+\log c $
……..(i) This passes through $ (2,1) $
$ -\log ( 1-\frac{1}{4} )=\log 2+\log c $
Therefore $ c=\frac{2}{3} $
From equation (i), $ \log ( \frac{x^{2}}{x^{2}-y^{2}} )=\log xc $
$ 2(x^{2}-y^{2})=3x $ .
BETA
