Solving ODE $yy'-y\cos x =1$ I am looking for some continuous $f$ that give rises to a non-monotonic solution to the differential equation
$$y\frac{dy}{dx}-yf(x) =1, \ \  x\in [0,\infty), \ \ y(0)=f(0)$$
Can you suggest some $f$ for which the solution is non-monotonic? 
Note that if $f$ itself always have same sign, then it forces $\frac{dy}{dx}$ to retain same sign. In that case solution is always monotonic. So I was searching for $f$ which oscillates. Something like $f(x)=\cos x$.
So I end up with the following ODE
$$y\frac{dy}{dx}-y\cos x =1, \ \  x\in [0,\infty), \ \ y(0)=1$$
I am not sure if it solvable at all. At least the techniques that I know doesn't apply to this problem. 
Any help and suggestions?
 A: Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=143:
Let $\begin{cases}x=-2\tan^{-1}4u\\y=r-\dfrac{8u}{16u^2+1}\end{cases}$ ,
Then $\dfrac{du}{dr}=-2ru^2+u-\dfrac{r}{8}$
Which reduces to a Riccati ODE.
Let $u=\dfrac{1}{2rv}\dfrac{dv}{dr}$ ,
Then $\dfrac{du}{dr}=\dfrac{1}{2rv}\dfrac{d^2v}{dr^2}-\dfrac{1}{2r^2v}\dfrac{dv}{dr}-\dfrac{1}{2rv^2}\left(\dfrac{dv}{dr}\right)^2$
$\therefore\dfrac{1}{2rv}\dfrac{d^2v}{dr^2}-\dfrac{1}{2r^2v}\dfrac{dv}{dr}-\dfrac{1}{2rv^2}\left(\dfrac{dv}{dr}\right)^2=-2r\left(\dfrac{1}{2rv}\dfrac{dv}{dr}\right)^2+\dfrac{1}{2rv}\dfrac{dv}{dr}-\dfrac{r}{8}$
$\dfrac{1}{2rv}\dfrac{d^2v}{dr^2}-\dfrac{1}{2r^2v}\dfrac{dv}{dr}-\dfrac{1}{2rv^2}\left(\dfrac{dv}{dr}\right)^2=-\dfrac{1}{2rv^2}\left(\dfrac{dv}{dr}\right)^2+\dfrac{1}{2rv}\dfrac{dv}{dr}-\dfrac{r}{8}$
$\dfrac{1}{2rv}\dfrac{d^2v}{dr^2}-\dfrac{1}{2r^2v}\dfrac{dv}{dr}-\dfrac{1}{2rv}\dfrac{dv}{dr}+\dfrac{r}{8}=0$
$\dfrac{d^2v}{dr^2}-\left(1+\dfrac{1}{r}\right)\dfrac{dv}{dr}+\dfrac{r^2v}{4}=0$
A: Although it is nice to know that this equation is a member of a named family, that does little to tell you the nature of its solution.
The observation from numeric work is that for $x > 100 \pi$, the following approximation is good to about a 0.01 absolute accuracy:
$$
y \approx \sqrt{2x} + \sin x + \frac1{\sqrt{2x}} - \frac12 \frac{\cos x}{x}
$$
Of these terms, only $\sin x$ has a derivative which is not small (on the scale of $\frac1{\sqrt{x}}$ or smaller, so this function definitely does oscillate and thus is a non-monotonic solution.
