A) \[\cos (x+y)+x=c\]
B) \[\cos (x+y)=c\]
C) \[\sin (x+y)+x=c\]
D) \[\sin (x+y)+\sin (x+y)=c\]
E) \[x-\cos (x+y)=c\]
Correct Answer: A
Solution :
\[\because \] \[\frac{dy}{dx}+1\cos ec(x+y)\] Let \[x+y=t\] \[\Rightarrow \] \[1+\frac{dy}{dx}=\frac{dt}{dx}\] \[\therefore \] \[\frac{dt}{\cos ect}=dx\] On integrating both sides, we get \[\int{\sin t\,dt}=\int{dx}\] \[\Rightarrow \] \[-\cos t=x+c\] \[\Rightarrow \] \[\cos (x+y)+x=c\]
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