Ask Questions, Get Answers

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

Solve the following $\cos^{2}xdy +ye^{\tan x }dx=0 $

Can you answer this question?

1 Answer

0 votes
  • First order , first degree DE
  • Variable separable : Variables of a DE are rearranged to separate then, ie
  • $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • Can be written as $ \large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
$\cos ^2 x dy+y e^{\tan x}dx=0$ divided by $y \cos ^2 x$
$\large\frac{dy}{y}+\large\frac{e^{\tan x}}{\cos^2 x}$$dx$$=0$
Step 2:
The variables have been seperated
$\int \large\frac{dy}{y} =- \int e^{\tan x} $$\sec^2 x dx +c$
$\log y=-e^{\tan x}+c$
answered Sep 4, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App