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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution $\large\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

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  • $1- \cos x = 2\large\frac{\sin^2x}{2}$ 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
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