Find all solutions of the differential equation
![\[ y' + y \cot x = 2 \cos x \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-510312ddfe1677677e1911828c16d824_l3.png "Rendered by QuickLaTeX.com")
on the interval 
. Prove that exactly one of these solutions is valid on the larger interval 
.
Proof. We apply Theorem 8.3 (page 310 of Apostol) with
![\[ P(x) = \cot x, \qquad Q(x) = 2 \cos x. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f9c09e2dad125725816d4a23a3ca6225_l3.png "Rendered by QuickLaTeX.com")
Further, we choose 
and obtain solutions in terms of 
, this gives us
![\[ A(x) = \int_{\frac{\pi}{2}}^x \cot t \, dt = \log (\sin x). \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8cccde13dfcd780f49ff175b991f5501_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\begin{align*} f(x) &= be^{- \log (\sin x)} + e^{-\log(\sin x)} \int_{\frac{\pi}{2}}^x 2(\cos t)e^{\log(\sin t)} \, dt \\[9pt] &= \frac{b}{\sin x} + \frac{1}{\sin x} \int_{\frac{\pi}{2}}^x \sin (2t) \, dt \\[9pt] &= \frac{b}{\sin x} - \frac{\cos (2x)}{2 \sin x} + \frac{1}{2 \sin x} \\[9pt] &= \frac{b}{\sin x} - \frac{1-2\sin^2 x}{2 \sin x} + \frac{1}{2 \sin x} \\[9pt] &= \frac{b}{\sin x} + \sin x. \end{align*}](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0c915c3bc33676f35d65719e6039f287_l3.png "Rendered by QuickLaTeX.com")
These are then all of the solutions valid on . The only one of these solutions valid on the interval is the one with 
, or 
The answer is slightly off: should be (b+1)/sinx + sinx , or C/sinx + sinx. cos(2*pi/2) is negative. So b=-1 ?
I’ve also found the following:
Same here