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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    Direction: (Q. Nos. 84) Let \[\frac{dy}{dx}=\frac{1}{x+y}\,\] and \[y(0)=0\]. Then
    At \[y={{\log }^{3}},\,\frac{dy}{dx}\] is equal to

    A)   1                                           

    B)  2            

    C)  \[\frac{1}{2}\]                                  

    D)  - 2

    Correct Answer: C

    Solution :

                    \[\frac{dx}{dy}-x=y\] \[\therefore \]  \[IF={{e}^{-\int{dy}}}={{e}^{-y}}\] \[\therefore \]  \[x{{e}^{-y}}=\int{y{{e}^{-y}}dy}\] \[\therefore \]  \[x{{e}^{-y}}\,=-(y+1){{e}^{-y}}+A\] \[\because \]     \[x=0,\,\,y=0\,\,\Rightarrow \,A=1\] \[x{{e}^{-y}}=-(y+1){{e}^{-y}}+1\] \[\Rightarrow \]               \[x={{e}^{y}}-(y+1)\] At           \[y=\log \,3,\] \[\Rightarrow \]               \[x=2-\,\log \,3\] \[\frac{dx}{dy}=2-\log \,3+\log \,3\] = 2 \[\therefore \]  \[\frac{dy}{dx}=\frac{1}{2}\]


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