Question

Solve the following differential equations
$(1+\cos x)dy=(1-\cos x)dx$

• A. $y=\tan \frac{x}{2}-x+c$
• B. $y=4\tan \frac{x}{2}-x+c$
• C. $y=2\tan \frac{x}{2}-x+c$
• D. None of these

Answer 1

The correct option is C $y=2\tan \frac{x}{2}-x+c$

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

$dy=\frac{1-\cos x}{1+\cos x}dx$

We know

$\cos 2x=2{\cos }^{2}x-1$

Putting $x=\frac{x}{2}$

$\cos 2.\frac{x}{2}=2{\cos }^2\frac{x}{2}-1$

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

$1+\cos x=2{\cos }^2\frac{x}{2}$

We know

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

Putting $x=\frac{x}{2}$

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

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

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

Putting the values in equation

$dy=\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}dx$

$dy=\frac{{\sin }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}}dx$

$dy={\tan }^2\frac{x}{2}dx$

We know that

${\tan }^{2}x+1={\sec }^{2}x$

${\tan }^{2}x={\sec }^{2}x-1$

Putting $x=\frac{x}{2}$

${\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}-1$

Now,

$dy=({\sec }^2\frac{x}{2}-1)dx$

Integrating both sides

$\int dy=\int ({\sec }^2\frac{x}{2}-1)dx$

$y=\int {\sec }^2\frac{x}{2}dx-\int dx$

$y=\frac{1}{\frac{1}{2}}\tan \frac{x}{2}-x+c$

$y=2\tan \frac{x}{2}-x+c$

This is the required general solution