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JEE Main & Advanced Mathematics Differential Equations Question Bank Mock Test - Differential Equations

  • 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

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

    B) \[2{{x}^{3}}-1\]

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

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

    Correct Answer: D

    Solution :

    [d] \[x(1-{{x}^{2}})dy+(2{{x}^{2}}y-y-a{{x}^{3}})dx=0\] Or \[x(1-{{x}^{2}})\frac{dy}{dx}+2{{x}^{2}}y-y-a{{x}^{3}}=0\] Or \[x(1-{{x}^{2}})\frac{dy}{dx}+y(2{{x}^{2}}-1)=a{{x}^{3}}\] Or \[\frac{dy}{dx}+\frac{2{{x}^{2}}-1}{x(1-{{x}^{2}})}y=\frac{a{{x}^{3}}}{x(1-{{x}^{2}})}\] Which is of the form \[\frac{dy}{dx}+Py=Q.\] Its integrating factor is\[{{e}^{\int{Pdx}}}\]. Here, \[P=\frac{2{{x}^{2}}-1}{x(1-{{x}^{2}})}\]


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