Differential Equations Question 285
Question: The equation of the curve satisfying $ xdy-ydx=\sqrt{x^{2}-y^{2}} $ and $ y(1)=0 $ is:
Options:
A) $ y=x^{2}\log (\sin x) $
B) $ y=x\sin (logx) $
C) $ y^{2}=x{{(x-1)}^{2}} $
D) $ y=2x^{2}(x-1) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The given equation can be rewritten as $ x^{2}[ \frac{xdy-ydx}{x^{2}} ]=x\sqrt{1-\frac{y^{2}}{x^{2}}} $
$ \Rightarrow x^{2}\frac{d}{dx}( \frac{y}{x} )=x\sqrt{1-\frac{y^{2}}{x^{2}}} $ Put $ \frac{y}{x}=z, $ we get $ {{\sin }^{-1}}(z)=logx+c\Rightarrow si{n^{-1}}( \frac{y}{x} )=\log x+c $ Apply the boundary value y(1) = 0
$ \Rightarrow {{\sin }^{-1}}( \frac{0}{1} )=\log 1+c\Rightarrow c=0 $
$ \therefore {{\sin }^{-1}}( \frac{y}{x} )=\log x+0\Rightarrow y=x\sin (logx) $
BETA
