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

  • question_answer
    Solution of the equation \[({{e}^{x}}+1)ydy=(y+1){{e}^{x}}dx\] is [AISSE 1988]

    A)                 \[c(y+1)({{e}^{x}}+1)+{{e}^{y}}=0\]     

    B)                 \[c(y+1)({{e}^{x}}-1)+{{e}^{y}}=0\]

    C)                 \[c(y+1)({{e}^{x}}-1)-{{e}^{y}}=0\]         

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

    Correct Answer: D

    Solution :

                       \[({{e}^{x}}+1)ydy=(y+1){{e}^{x}}dx\]                    Þ \[\left( \frac{y}{y+1} \right)dy=\left( \frac{{{e}^{x}}}{{{e}^{x}}+1} \right)\,dx\]Þ \[\left[ 1-\frac{1}{y+1} \right]dy=\left( \frac{{{e}^{x}}}{{{e}^{x}}+1} \right)\,dx\]                    Þ \[\int_{{}}^{{}}{\left\{ 1-\frac{1}{y+1} \right\}}dy=\int_{{}}^{{}}{\frac{{{e}^{x}}}{{{e}^{x}}+1}}dx\]                                 Þ \[y=\log (y+1)+\log ({{e}^{x}}+1)+\log c\]or \[{{e}^{y}}=c(y+1)({{e}^{x}}+1)\].


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