Differential Equations Question 188
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
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 thee
Show Answer
Answer:
Correct Answer: C
Solution:
[c] 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} $ Or $ {{\sec }^{2}}udu=-\frac{1}{x}dx $
On integration, we get $ \tan u=-\log x+\log C $ Or $ \tan ( \frac{y}{x} )=-\log x+\log C $ This passes through $ (1,\pi /4). $ Therefore, $ 1=\log C. $ So, $ \tan ( \frac{y}{x} )=-\log x+1 $
$ =-\log x+{\log _{e}} $ Or $ y=x{{\tan }^{-1}}( \log ( \frac{e}{x} ) ) $
BETA
