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JEE Main & Advanced Mathematics Differential Equations Question Bank Variable Separable type differential equations

  • question_answer
    The solution of the differential equation \[\frac{dy}{dx}+\frac{1+{{x}^{2}}}{x}=0\] is 

    A)                 \[y=-\frac{1}{2}{{\tan }^{-1}}x+c\]              

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

    C)                 \[y=\frac{1}{2}{{\tan }^{-1}}x+c\]

    D)                 \[y-\log x-\frac{{{x}^{2}}}{2}=c\]

    Correct Answer: B

    Solution :

                       \[\frac{dy}{dx}+\frac{1+{{x}^{2}}}{x}=0\] Þ \[dy+\left( \frac{1}{x}+x \right)\text{ }dx=0\]                                 On integrating, we get \[y+\log x+\frac{{{x}^{2}}}{2}+c=0\].


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