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 :
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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