MHT CET · Maths · Differential Equations
The general solution of the differential equation \(\sec ^{2} x\) tany \(d x+\sec ^{2} y \tan x d y=0\)
is
- A \(\tan x \tan y=c\)
- B \(\sec x \tan y=c\)
- C \(\sec x \sec y=c\)
- D \(\tan x \sec y=c\)
Answer & Solution
Correct Answer
(A) \(\tan x \tan y=c\)
Step-by-step Solution
Detailed explanation
\(\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0 \)
\( \therefore \sec ^{2} x \tan y d x=-\sec ^{2} y \tan x d y \)
\( \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y \Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\) \(\int \frac{\sec ^{2} y}{\tan y} d y \)
\( \log (\tan x)=-\log (\tan y)+\log c \)
\( \therefore \log (\tan x \tan y)=\log c \)
\( \tan x \tan y=c\)
\( \therefore \sec ^{2} x \tan y d x=-\sec ^{2} y \tan x d y \)
\( \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y \Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\) \(\int \frac{\sec ^{2} y}{\tan y} d y \)
\( \log (\tan x)=-\log (\tan y)+\log c \)
\( \therefore \log (\tan x \tan y)=\log c \)
\( \tan x \tan y=c\)
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