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If y=y(x) is the solution of the differential equation \frac{5+e^{x}}{2+y}.\frac{dy}{dx}+e^{x}=0 satisfying y(0)=1, then a value of y\left ( log_{e}13 \right ) is :
Option: 1 1
Option: 2 -1
Option: 3 0
Option: 4 2

Answers (1)

best_answer

\\ \frac{1}{2+y} \frac{d y}{d x}=\frac{-e^{x}}{5+e^{x}} \\ \Rightarrow \frac{d y}{2+y}=\frac{-e^{x}}{5+e^{x}} \cdot d x \\\Rightarrow \int\frac{d y}{2+y}=\int\frac{-e^{x}}{5+e^{x}} \cdot d x

\\\ln (y+2)=-\ln \left(e^{x}+5\right)+\ln C\\\mathrm{\;\;\;\;\;\;}\;\;\;\;\;\;\;\;=\ln\left (\frac{C}{e^x+5} \right )\\\Rightarrow 2+y=\frac{C}{e^x+5}\\\Rightarrow y=\frac{C}{e^x+5}-2

y(0) = 1   (Given)

\\\Rightarrow 1=\frac{C}{e^0+5}-2\\\Rightarrow C=18

\\\Rightarrow y=\frac{18}{e^{\log_e13}+5}-2\\\Rightarrow y=-1

Posted by

himanshu.meshram

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