A) \[y=\tan \,\left( \frac{x+y}{2} \right)+c\]
B) \[y+{{\cos }^{-1}}\left( \frac{y}{x} \right)=c\]
C) \[y=x\,\,\sec \left( \frac{y}{x} \right)+c\]
D) None of these
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), \[\cos v\,\left( \frac{dv}{dx}-1 \right)=1\] Þ \[\cos v\,\frac{dv}{dx}=1+\cos v\] Þ \[\frac{\cos v}{1+\cos v}dv=dx\] Þ\[\left[ \frac{2{{\cos }^{2}}(v/2)-1}{2{{\cos }^{2}}(v/2)} \right]\,dv=dx\]Þ\[\left[ 1-\frac{1}{2}{{\sec }^{2}}(v/2) \right]\,dv=dx\] Integrate, \[v-\tan (v/2)=x+c\] \[x+y-\tan \left( \frac{x+y}{2} \right)=x+c\] Þ \[y=\tan \left( \frac{x+y}{2} \right)+c\].
You need to login to perform this action.
You will be redirected in
3 sec
