A) \[-2\]
B) \[-1\]
C) \[0\]
D) \[1\]
Correct Answer: B
Solution :
[b] \[(2xy-{{y}^{2}}-y)dx=(2xy+x-{{x}^{2}})dy\] \[\Rightarrow \,\,\,\,(2xydx+{{x}^{2}}dy)=(xdy+ydx)+(2xydy+{{y}^{2}}dx)\]\[\Rightarrow \,\,\,d({{x}^{2}}y)=d(xy)+d{{(xy)}^{2}}\] \[\Rightarrow \,\,\,{{x}^{2}}y=xy+x{{y}^{2}}+c\] At \[x=1,\] \[y=1,\] we get \[c=-1\] \[\Rightarrow \,\,\,\,{{x}^{2}}y=xy+x{{y}^{2}}-1\] Putting \[x=-1,\]we get \[y=-y+{{y}^{2}}-1\] \[\Rightarrow \,\,\,\,\,{{y}^{2}}+2y+1=0\] \[\Rightarrow \,\,\,\,y=-1\]You need to login to perform this action.
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