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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    The general solution of the differential equation \[\left( {{y}^{2}}{{x}^{3}} \right)dxxydy=0(x\ne 0)\] is:
    (where c is a constant of integration)
    [JEE Main 12-4-2019 Afternoon]

    A) \[{{y}^{2}}+2{{x}^{3}}+c{{x}^{2}}=0\]

    B) \[{{y}^{2}}+2{{x}^{2}}+c{{x}^{3}}=0\]

    C) \[{{y}^{2}}2{{x}^{3}}+c{{x}^{2}}=0\]

    D) \[{{y}^{2}}2{{x}^{2}}+c{{x}^{3}}=0\]

    Correct Answer: A

    Solution :

    \[xy\frac{dy}{dx}-{{y}^{2}}+{{x}^{3}}=0\]             put\[{{y}^{2}}=k\Rightarrow y\frac{dy}{dx}=\frac{1}{2}\frac{dk}{dx}\] \[\therefore \]given differential equation becomes \[\frac{dk}{dx}+k\left( -\frac{2}{x} \right)=-2{{x}^{2}}\] I.F.\[={{e}^{\int_{{}}^{{}}{-\frac{2}{x}dx}}}=\frac{1}{{{x}^{2}}}\] \[\therefore \]solution is \[k.\frac{1}{{{x}^{2}}}=\int_{{}}^{{}}{-2{{x}^{2}}.\frac{1}{{{x}^{2}}}dx+\lambda }\] \[{{y}^{2}}+2{{x}^{3}}=\lambda {{x}^{2}}\] take \[\lambda =-c\](integration constant)


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