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  • Solution of the differential equation t any sec2x dx + tanx sec2 ydy = 0 is: A. tanx + tany = k B. tanx – tan y = k C. D. tanx . tany = k

Solution of the differential equation  \tan y\sec^2x dx + \tan x \sec^2 ydy = 0 is:
A. tanx + tany = k
B. tanx – tan y = k
C. \frac{\tan x}{\tan y}= k
D. tanx . tany = k

Answers (1)

The given differential equation is
\tan y\sec^2x dx + \tan x \sec^2 ydy = 0
Divide it by tanx tany
$$ \Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\operatorname{tany}} d y=0 $$
Integrate
$$ \Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=0 $$
Put tanx=t hence,
\sec ^{2} x d x=d t $$
Put tany =z  hence
$$ \frac{d z}{d y}=\sec ^{2} y $$
That is \sec ^{2} y d y=d t$
\Rightarrow \int \frac{d t}{t}+\int \frac{d z}{z}=0$

\log t+\log z+c=0$
Resubstitue t and z

\log (\tan x)+\log (\tan y)+c=0$
$using \log a+\log \mathrm{b}=\log \mathrm{ab}$
\log ($ tanxtany $)=-\mathrm{c}$
\tan x \tan y $=e^{-c}$

e^{-c}$is constant because e is a constant and c is the integration constant let it be denoted as \mathrm{e}^{-c}=\mathrm{k}\\

\tan x \tan y $=\mathrm{k}$

Option D is correct.

Posted by

infoexpert22

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