Differential Equations Question 351
Question: The solution of the equation $ {{\sin }^{-1}}( \frac{dy}{dx} )=x+y $ is
Options:
A) $ \tan (x+y)+\sec (x+y)=x+c + \sec (x+y) \tan (x+y) $
B) $ \tan (x+y)-\sec (x+y)=x+c $
C) $ \tan (x+y)+\sec (x+y)+x+c=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ \frac{dy}{dx}=\sin (x+y) $
Now put $ \log (1+y)=x+\frac{x^{2}}{2}+\log c $ and $ \frac{dy}{dx}=\frac{dv}{dx} $
Therefore $ \frac{dy}{dx}=\sin (x+y) $ reduces to $ \frac{dv}{dx}=1+\sin v $
Now on integrating both sides, we get $ \tan v-\sec v=x+c $ or $ \tan (x+y)-\sec (x+y)=x+c $ .
BETA
