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

  • question_answer
    The solution of the differential equation \[3{{e}^{x}}\tan ydx+(1-{{e}^{x}}){{\sec }^{2}}ydy=0\] is [MP PET 1993; AISSE 1985]

    A)                 \[\tan y=c{{(1-{{e}^{x}})}^{3}}\]

    B)                 \[{{(1-{{e}^{x}})}^{3}}\tan y=c\]

    C)                 \[\tan y=c(1-{{e}^{x}})\]  

    D)                 \[(1-{{e}^{x}})\tan y=c\]

    Correct Answer: A

    Solution :

                       It can be written in the form of         \[\frac{{{\sec }^{2}}y}{\tan y}dy=-3\frac{{{e}^{x}}}{1-{{e}^{x}}}dx\]         \[\int{\frac{{{\sec }^{2}}y}{\tan y}}dy=-3\int{\frac{{{e}^{x}}}{1-{{e}^{x}}}dx}\]                 Þ \[\log (\tan y)=3\log (1-{{e}^{x}})+\log c\] Þ \[\tan y=c{{(1-{{e}^{x}})}^{3}}\].


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