Differential-Equations Question 372
Question: The solution of $ \cos (x+y),dy=dx $ is
[DCE 1999]
Options:
A) $ y=\tan ,( \frac{x+y}{2} )+c $
B) $ y+{{\cos }^{-1}}( \frac{y}{x} )=c $
C) $ y=x\sec ( \frac{y}{x} )+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \cos (x+y)dy=dx $ …..(i) Put $ x+y=v $ . Differentiate $ 1+\frac{dy}{dx}=\frac{dv}{dx} $ Put these values in (i), $ \cos v,( \frac{dv}{dx}-1 )=1 $
Þ $ \cos v,\frac{dv}{dx}=1+\cos v $
Þ $ \frac{\cos v}{1+\cos v}dv=dx $
Þ $ [ \frac{2{{\cos }^{2}}(v/2)-1}{2{{\cos }^{2}}(v/2)} ],dv=dx $
Þ $ [ 1-\frac{1}{2}{{\sec }^{2}}(v/2) ],dv=dx $ Integrate, $ v-\tan (v/2)=x+c $ $ x+y-\tan ( \frac{x+y}{2} )=x+c $
Þ $ y=\tan ( \frac{x+y}{2} )+c $ .
BETA
