Differential Equations Question 10
Question: The solution of $ \frac{dy}{dx}=\sin (x+y)+\cos (x+y) $ is
Options:
A) $ \log [ 1+\tan ( \frac{x+y}{2} ) ]+c=0 $
B) $ \log [ 1+\tan ( \frac{x+y}{2} ) ]=x+c $
C) $ \log [ 1-\tan ( \frac{x+y}{2} ) ]=x+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ x+y=v $ and $ 1+\frac{dy}{dx}=\frac{dv}{dx} $
Therefore, the differential equation reduces to $ \frac{dv}{dx}=(1+\cos v)+\sin v $
$ =2{{\cos }^{2}}\frac{v}{2}+2\sin \frac{v}{2}\cos \frac{v}{2}=2{{\cos }^{2}}\frac{v}{2}( 1+\tan \frac{v}{2} ) $
Therefore $ \int _{{}}^{{}}{\frac{{{\sec }^{2}}(v/2)dv}{2[1+\tan (v/2)]}}=\int _{{}}^{{}}{dx} $
Therefore $ \log [ 1+\tan ( \frac{x+y}{2} ) ]=x+c $ .
BETA
