logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

General solution of differential equation : $x^2e^y dy - (x-1) e^x dx=0$

$(a)\;e^x=\frac{e^y}{x}+c \\ (b)\;e^y=\frac{e^x}{x}+c \\ (c)\;e^x= \frac{e^y}{y}+c \\ (d)\;e^y= \frac{e^x}{y}+c $
Can you answer this question?
 
 

1 Answer

0 votes
$x^2e^y dy -(x-1) ^x edx=0$
$e^y dy =\large\frac{x-1}{x^2}$$e^x dx$
$d(e^y) =e^x \bigg[ \large\frac{1}{x}- \frac{1}{x^2}\bigg]$$dx$
$d(e^y)= d \bigg( \frac{e^x}{x}\bigg)$
$e^y =\large\frac{e^x}{x}$$+c$
Hence b is the correct answer.
answered Feb 3, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...