Medium MCQ +4 / -1 PYQ · JEE Mains 2020

If y = y(x) is the solution of the differential

equation ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ satisfying y(0) = 1, then a value of y(loge13) is :

  1. A -1 Correct answer
  2. B 1
  3. C 0
  4. D 2

Solution

${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ <br><br>$\Rightarrow$ ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x}$ <br><br>Integrating both sides, <br><br>$\Rightarrow$ $\int {{{dy} \over {2 + y}}} = \int {{{ - {e^x}} \over {{e^x} + 5}}} dx$ <br><br>$\Rightarrow$ ln (y + 2) = – ln(e<sup>x</sup> + 5) + k <br><br>$\Rightarrow$ (y + 2) (e<sup>x</sup> + 5) = C <br><br>$\because$ y(0) = 1 <br><br>$\Rightarrow$ C = 18 <br><br>$\therefore$ y + 2 = ${{{18} \over {{e^x} + 5}}}$ <br><br>At x = log<sub>e</sub>13 <br><br>y + 2 = ${{{18} \over {13 + 5}}}$ = 1 <br><br>$\Rightarrow$ y = -1

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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