JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Let y= y(x) be the solution curve of the differential equation, \[({{y}^{2}}-x)\frac{dx}{dx}=1,\] satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is [JEE MAIN Held on 07-01-2020 Evening]

    A) 2 - e

    B) 2 + e

    C) - e                   

    D) 2

    Correct Answer: A

    Solution :

    [a] \[\because \,\,\,({{y}^{2}}-x)\frac{dy}{dx}=1\] \[\therefore \,\,\,\,\,{{y}^{2}}dy-xdy=dx\] \[\Rightarrow \,\,{{e}^{y}}\cdot {{y}^{2}}dy={{e}^{y}}dx+x{{e}^{y}}dy\] \[\Rightarrow \,\,{{e}^{y}}\cdot {{y}^{2}}dy=d({{e}^{y}}\cdot x)\] On integrating both sides we get \[\int{{{e}^{y}}\cdot {{y}^{2}}dy=\int{d({{e}^{y}}\cdot x)}}\] \[{{y}^{2}}\cdot {{e}^{y}}-\int{2y\cdot {{e}^{y}}dy}={{e}^{y}}\cdot x\] \[{{y}^{2}}\cdot {{e}^{y}}-2\left\{ y\cdot {{e}^{y}}-\int{{{e}^{y}}dy} \right\}={{e}^{y}}\cdot x\] \[\therefore \,\,\,{{y}^{2}}\cdot {{e}^{y}}-2y{{e}^{y}}+2{{e}^{y}}+2{{e}^{y}}={{e}^{y}}\cdot x+c\] \[\therefore \,\,\,\,\,y(0)=1\] \[\Rightarrow \,\,c=e\] \[\therefore \,\,{{y}^{2}}-2y+2=x+e\cdot {{e}^{-y}}\] \[\therefore \]    when y = 0 then x = 2 - e      

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