A) \[{{x}^{2}}+{{y}^{2}}=1\]
B) \[{{x}^{2}}-{{y}^{2}}=1\]
C) \[2{{x}^{2}}+{{y}^{2}}=2\]
D) None of these
Correct Answer: B
Solution :
Given, differential equation is \[(1+{{y}^{2}})dx-xydy=0\] \[\Rightarrow \] \[(1+{{y}^{2}})dx=xy\,dy\] \[\Rightarrow \] \[\frac{dx}{x}=\frac{y\,dy}{1+{{y}^{2}}}\] On integrating both sides, \[\log x=\frac{1}{2}\log (1+{{y}^{2}})+\log c\] \[\Rightarrow \] \[\log x=\log \sqrt{1+{{y}^{2}}}.c\] \[\Rightarrow \] \[x=c\sqrt{1+{{y}^{2}}}\] At point (1, 0), \[1=c\sqrt{1+0}\] \[\Rightarrow \] \[c=1\] \[\therefore \] \[x=\sqrt{1+{{y}^{2}}}\] On squaring both the sides, we get \[{{x}^{2}}-{{y}^{2}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
