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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 26

  • question_answer
    If \[(2xy-{{y}^{2}}-y)dx=92xy+x-{{x}^{2}})dy\] and \[y(1)=1,\] then the value of \[y(-1)\] is

    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 \[y=-y+{{y}^{2}}-1\] \[\Rightarrow \,\,\,\,\,{{y}^{2}}+2y+1=0\] \[\Rightarrow \,\,\,\,y=-1\]    


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