A) \[\frac{2x-1}{2y+3}=c\]
B) \[\frac{2x+3}{2y-1}=c\]
C) \[\frac{2x-3}{2y-1}=c\]
D) \[\frac{2y+1}{2x-3}=c\]
E) \[\frac{2x+1}{2y-3}=c\]
Correct Answer: B
Solution :
The given equation is \[(2y-1)dx-(2x+3)dy=0\] Or it can be rewritten as \[\frac{dx}{(2x+3)}-\frac{dy}{(2y-1)}=0\] On integrating both sides \[\Rightarrow \] \[\int{\frac{1}{(2x+3)}}dx-\int{\frac{1}{(2y-1)}}dy=0\] \[\Rightarrow \] \[\frac{1}{2}\log (2x+3)-\frac{1}{2}log(2y-1)=\log c\] \[\Rightarrow \] \[\log \left( \frac{2x+3}{2y-1} \right)=\log c\] \[\Rightarrow \] \[\frac{2x+3}{2y-1}=c\]
You need to login to perform this action.
You will be redirected in
3 sec
