Differential Equations Question 330
Question: The slope of the tangent at $ (x,y) $ to a curve passing through $ ( 1,\frac{\pi }{4} ) $ is given by $ \frac{y}{x}-{{\cos }^{2}}( \frac{y}{x} ) $ , then the equation of the curve is
[Kurukshetra CEE 2002]
Options:
A) $ y={{\tan }^{-1}}[ \log ( \frac{e}{x} ) ] $
B) $ y=x{{\tan }^{-1}}[ \log ( \frac{x}{e} ) ] $
C) $ y=x{{\tan }^{-1}}[ \log ( \frac{e}{x} ) ] $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ \frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}( \frac{y}{x} ) $
Putting $ y=vx $ so that $ \frac{dy}{dx}=v+x\frac{dv}{dx} $ , we get
$ v+x\frac{dv}{dx}=v-{{\cos }^{2}}v $ or $ \frac{dv}{{{\cos }^{2}}v}=-\frac{dx}{x} $
On integrating, we get $ \tan v=-\log x+\log c $
Therefore $ \tan ( \frac{y}{x} )=-\log x+\log C $
This passes through $ ( 1,\frac{\pi }{4} ) $ , therefore $ 1=\log c $
Therefore $ \tan ( \frac{y}{x} )=-\log x+\log e $
Therefore $ y=x{{\tan }^{-1}}[ \log ( \frac{e}{x} ) ] $ .
BETA
