Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Find the general solution $\large\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

Can you answer this question?

1 Answer

0 votes
  • $1- \cos x = 2\large\frac{\sin^2x}{2}$ and $1 + \cos x = 2\large\frac{\cos^2x}{2}$
Step 1:
Given: $\large\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$
Using the hint from the tool box
Step 2:
Seperating the variables we get,
$dy = \large\frac{\tan^2x}{2} $$dx$
Integrating on both sides we get
$\int dy=\int \tan^2\large\frac{x}{2}$$dx$
But $\tan^2\large\frac{x}{2} = $$\sec^2\large\frac{x}{2}$$-1$
substituting this we get
$y = \int (\sec^2\large\frac{x}{2}$$ - 1) dx$
This is the required general solution.
answered Jul 31, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App