# Find the particular solution of the differential equationgiven that when x = 0.

Answers (1)

$\\ \cos y d x=-\left(1+ e^{-x}\right) \sin y d y \\\\ \frac{d x}{\left(1+ e^{-x}\right)}=-\frac{\sin y}{\cos y} d y \\\\ \frac{d x}{\left(1+ e^{-x}\right)}=-\tan y d y \\\\ \frac{d x}{\left(1+\frac{1}{e^{x}}\right)}=-\tan y dy \\\\ \frac{d x}{\left(\frac{e^{x}+1}{e^{x}}\right)}=-\tan y d y$

$\\ \frac{e^{x} d x}{\left(e^{x}+1\right)}=-\tan y \mathrm{d} y \\\\ Integrating both sides \\\\ \int \frac{e^{x} d x}{\left(e^{x}+1\right)}=-\int \tan y d y \\\\ \int \frac{e^{x} d x}{\left(e^{x}+1\right)}=-\log |\sec y|+C$

$\\ Let e^{x}+1=t \\ \\ e^{x} d x=d t \\\\ \int \frac{d t}{t}=-\log |\sec y|+C \\ \\ \log |t|=-\log |\sec y|+C \\\\ Put value of t$

$\\ \log \left|e^{x}+1\right|=-\log |\sec y|+C \\\\ \log \left|e^{x}+1\right|+\log |\sec y|=C \\\\ \because \log a+ \log b=\log ab \\\\ \log \left|\left(e^{x}+1\right) \times \sec y\right|=C \\\\ \log \left|\left(e^{x}+1\right) \times \sec y\right|=\log k \\\\ \left(e^{x}+1\right) \times \sec y= k \\\\ \left(e^{x}+1\right) \times \frac{1}{\cos y}=k \\\\ \left(e^{x}+1\right)=k\cos \mathrm{y}$

At  x = 0 ,

$\\ \left(e^{0}+1\right)=k\cos \frac{\pi}{4}$

$k= 2 \sqrt{2}$

The particular solution:

$\\ \left(e^{x}+1\right)=2 \sqrt{2}\cos y$

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