Behavior of solutions of $y'+p(x)y=p(x)\sin(\frac{1}{x+1})$ Consider the linear d.e $$(E) \ \ \ \ \ y'+p(x)y=p(x)\sin(\frac{1}{x+1}), x\geq 0$$
which $p(x)=(x+1)(x-1)(x-2), x\geq 0$. I want to study the behavior of solutions of $(E)$.
Specifically:

*

*Does every solution of $(E)$ tend to $0$, when $x\to +\infty$?

*Does the equation $(E)$ have oscillating solutions?

*Does every solution of $(E)$ finally positive?

About question 1:
I showed that $p(x)$ is finally lower bounded by a positive number (it's easy to show it) and then I calculate $$\lim_{x\to+\infty} p(x)\sin(\frac{1}{x+1}) $$ which is (unfortunately) equals to $+\infty$ and not $0$ as I hoped, before this calculation.
About question 2:
I have no Idea how exactly can I find (or not) a solution with infinite countable  roots.
Any hint would be helpful, please. I'm stuck
 A: About question 1:
Every solution is given by the form
$$y(x)= e^{-P(x)}\left(y(0)-\sin 1 +\sin (\frac{1}{x+1})+\int_0^x \frac{e^{P(t)}}{(1+t)^2}  \cos (\frac{1}{1+t}) dt\right),\ x\geq 0.$$
I observe that there exists k>0 such that $p(x)>k, \forall x\geq 3/2.$
Consequently, $e^{-P(x)}\to 0$, when $\  x\to +\infty$. Thus,  $e^{-P(x)}(y(0)-\sin 1 +\sin (\frac{1}{x+1}))\to 0$, when $x\to +\infty$. I consider now the function
$$F(x)=\int_0^x \frac{e^{P(t)}}{(1+t)^2}  \cos (\frac{1}{1+t}) dt,\ x\geq 0$$
For all $x\geq 0$, we have
$$0\leq |F(x)|\leq  e^{-P(x)}\int_0^x \frac{e^{P(t)}}{(1+t)^2} dt $$
Let $G(x)=\int_0^x \frac{e^{P(t)}}{(1+t)^2} dt, \ x\geq 0$. I observe that $G$ is an increasing function, as a result $G(x)\to l\geq 0$ or $G(x)\to +\infty$, as $x\to \infty$. In first case (the second is obvious) by L'Hospital rule we get that $|F(x)|\to 0,$ as $ x\to \infty$.
Consequently, every solution of $(Ε)$ trends to 0, as $x$ tends to $+\infty$.
A: Multiply both members by $e^{P(x)}$ where $P(x)=\int_0^x p(t) dt$. Then
$(e^{P(x)}y)’=(e^{P(x)})’sin(\frac{1}{x+1})$
Integrate both members wrt $x$ to get
$$e^{P(x)}y-y(0)= e^{P(x)} sin(\frac{1}{x+1})-sin(1)+\int_0^x \frac{1}{(t+1)^2}cos(\frac{1}{t+1})e^{P(t)}dt$$
$$y=(y(0)-sin(1)+  \int_0^x \frac{1}{(t+1)^2}cos(\frac{1}{t+1})e^{P(t)}dt )e^{-P(x)}+ sin(\frac{1}{x+1})$$
(1) You can use de l’Hopital at $\infty$ to prove that $y$ go to $0$ for $x\mapsto +\infty$, if the coefficient $a_{\deg(P(x))}$ of the polynomial $P(x)$ is strictly positive, so that of $p(x)$ is strictly positive;
(2) Now you can modify the  solution of your ODE in order to understand the behaviour, in the following way
$S(x):= \int_0^x \frac{1}{(t+1)^2}cos(\frac{1}{t+1})e^{P(t)}dt$
If you impose the change variable $z:= \frac{1}{1+t}$ you get
$$S(x)=-\int_1^{\frac{1}{1+x}} cos(z)e^{P(\frac{1-z}{z}) }dz= \int_{\frac{1}{1+x}}^1 cos(z)e^{P(\frac{1-z}{z}) }dz$$
that is an increasing function for $x\geq 0$, because $cos(z)e^{P(\frac{1-z}{z})}$ is positive and the integration interval is bigger for $x_2\geq x_1$. Call  $T(u):=S(\frac{1-u}{u})= \int_{u}^1 cos(z)e^{P(\frac{1-z}{z}) }dz$ that is a function on $u\in (0,1]$.
The next step is to change variable for your solution $y$:
$u:= \frac{1}{1+x}$
In this case
$y(u)= \left( y(0)-sin(1)+ T(u) \right)e^{-P(\frac{1-u}{u})}+ sin(u)$
with $u\in (0,1]$
But now you have done because if $y$ is an oscillating function, then $y’$ has to be an oscillating function, by Rollè theorem. But now
$$y’(u)= \frac{-cos(u)e^{2P(\frac{1-u}{u})}+P’(\frac{1-u}{u}) \left( y(0)-sin(1)+ T(u) \right) e^{P(\frac{1-u}{u})}}{e^{2P(\frac{1-u}{u})}}+cos(u)$$
so $y’(u)=0$ if and only if $P’(\frac{1-u}{u}) \left( y(0)-sin(1)+ T(u) \right) =0$
Now $T(u)$ is a positive decreasing function, so $ y(0)-sin(1)+ T(u) $ has exactly $1$ zero, when $y(0)\leq sin(1)$, otherwise has not zeros. Meanwhile $P’=-\frac{1}{u^2}p(\frac{1-u}{u})$ has at most $\deg(p)$ zeros.
(3) Suppose $y(0)\geq sin(1)$. Then $y(u)\geq 0$ for each $u\in (0,1]$, so that $y(x)$ is always positive for each $x\geq 0$.
Conversely suppose $y(0)\leq sin(1)$. Let $u_0$ be the only root of $ y(0)-sin(1)+ T(u) $. Then for $u\leq u_0$, $T(u)\geq T(u_0)$ and so
$$y(0)-sin(1)+ T(u)= (y(0)-sin(1)+ T(u_0))+T(u)-T(u_0)=T(u)-T(u_0)\geq 0$$ and so $y(u)\geq 0$ for each $u\leq u_0$.
This means $y(x)\geq 0$ for each $x\geq \frac{1-u_0}{u_0}$ so that $y$ is finally positive.
Thus at the end of the story properties  (1), (2) and (3) hold.
