A) \[-{{\log }_{e}}\left| \frac{1+x-y}{1-x+y} \right|=x+y-2\]
B) \[{{\log }_{e}}\left| \frac{2-x}{2-y} \right|=x-y\]
C) \[-{{\log }_{e}}\left| \frac{1-x+y}{1+x-y} \right|=2(x-1)\]
D) \[{{\log }_{e}}\left| \frac{2-y}{x-x} \right|=2(y-1)\]
Correct Answer: C
Solution :
Given,\[\frac{dy}{dx}={{(x-y)}^{2}}\] Let\[x-y=t\Rightarrow \frac{dy}{dx}=1-\frac{dt}{dx}\] \[\therefore \]\[1-\frac{dt}{dx}={{t}^{2}}\] \[\Rightarrow \]\[\int_{{}}^{{}}{\frac{dt}{1-{{t}^{2}}}}=\int_{{}}^{{}}{dx}\Rightarrow \frac{1}{2}{{\log }_{e}}\left| \frac{1+t}{1-t} \right|=x+c\] \[\Rightarrow \]\[\frac{1}{2}{{\log }_{e}}\left| \frac{1+x-y}{1-x+y} \right|=x+C\] Given,\[y(1)=1\] \[\therefore \]\[\frac{1}{2}{{\log }_{e}}(1)=C+1\Rightarrow C=-1\] \[\therefore \]\[{{\log }_{e}}\left| \frac{1+x-y}{1-x+y} \right|=2(x-1)\] \[\Rightarrow \]\[-{{\log }_{e}}\left| \frac{1-x+y}{1+x-y} \right|=2(x-1)\]
You need to login to perform this action.
You will be redirected in
3 sec
