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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    If integrating factor of \[x(1-{{x}^{2}})dy+(2{{x}^{2}}y-y-a{{x}^{3}})dx=0\] is \[{{e}^{\int_{{}}^{{}}{Pdx}}},\] then P is equal to [MP PET 1999]

    A) \[\frac{2{{x}^{2}}-a{{x}^{3}}}{x(1-{{x}^{2}})}\]       

    B) \[(2{{x}^{2}}-1)\]

    C) \[\frac{2{{x}^{2}}-1}{a{{x}^{3}}}\]                                 

    D) \[\frac{(2{{x}^{2}}-1)}{x(1-{{x}^{2}})}\]

    Correct Answer: D

    Solution :

    • \[x(1-{{x}^{2}})dy+(2{{x}^{2}}y-y-a{{x}^{3}})dx=0\]                   
    • \[\frac{dy}{dx}+\frac{(2{{x}^{2}}-1)}{x(1-{{x}^{2}})}y=\frac{a{{x}^{2}}}{(1-{{x}^{2}})}\],  \[\therefore P=\frac{2{{x}^{2}}-1}{x(1-{{x}^{2}})}\].


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