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  • Find the particular solution of the differential equation<img alt="e^{x}\tan ydx+\left ( 2-e^{x} \right )\sec ^{2}y\, dy= 0," src="https://entrancecorner.codecogs.com/gif.latex?e%5E%7Bx%7D%5Ctan%20ydx&plus;%5Cleft%20%28%202-e%5E%7Bx%7D%20%5Cright%2

Find the particular solution of the differential equatione^{x}\tan ydx+\left ( 2-e^{x} \right )\sec ^{2}y\, dy= 0, given that y= \frac{\pi }{4}  when x= 0\cdot

 

 

 

 
 
 
 
 

Answers (1)

given differential equation is
e^{x}\tan ydx+\left ( 2-e^{x} \right )\sec ^{2}ydy= 0
\Rightarrow e^{x}\tan ydx= \left ( e^{x}-2 \right )\sec ^{2}ydy
\Rightarrow \frac{e^{x}}{ e^{x}-2}dx= \frac{\sec ^{2}y}{\tan y}dy
Integrating both sides, we get
\int \frac{e^{x}}{ e^{x}-2}dx=\int \frac{\sec ^{2}y}{\tan y}dy
\Rightarrow \log \left | e^{x}-2 \right |= \log \left | \tan y \right |+\log c
\Rightarrow \log \left | e^{x}-2 \right |= \log \left |c \tan y \right |
\Rightarrow e^{x}-2= c\tan y---(i)
put y= \frac{\pi }{4}\; \; when\, x= 0
e^{0}-2= c\left ( \tan \frac{\pi }{4} \right )
\Rightarrow -1= c\;, \; c= -1
\Rightarrow\therefore From equation (i)
e^{x}-2=- \tan y\Rightarrow y= \tan^{-1}\left ( 2-e^{x} \right ) which is the required solution

Posted by

Ravindra Pindel

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