A) \[{{(y-1)}^{2}}(1-{{x}^{2}})=0\]
B) \[{{(y-1)}^{2}}{{(1-x)}^{2}}={{c}^{2}}{{y}^{2}}\]
C) \[{{(y-1)}^{2}}(1+{{x}^{2}})={{c}^{2}}{{y}^{2}}\]
D) None of these
Correct Answer: B
Solution :
\[(1-{{x}^{2}})dy+xydx=x{{y}^{2}}dx\] Þ \[(1-{{x}^{2}})dy=xy(y-1)dx\] Þ \[\frac{1}{y(y-1)}dy=\frac{x}{(1-{{x}^{2}})}dx\] Now on integrating both sides, we get \[\log (y-1)-\log y=-\frac{1}{2}\log (1-{{x}^{2}})+\log c\] or \[2\log (y-1)+\log (1-{{x}^{2}})=\log {{y}^{2}}{{c}^{2}}\] Hence the solution is \[{{(y-1)}^{2}}(1-{{x}^{2}})={{c}^{2}}{{y}^{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
