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We are given the differential equation $\large\frac{\textrm dy}{\textrm dx}(x)=\frac{1}{3y(x)+x^2}+\frac{2}{4+x^2}$ and its directional field (see image below).

If this equation is solvable in closed form, I suspect it is not an easy solution and Wolfram Alpha confirms this idea by not giving such a solution.

We are asked:

  • $\lim\limits_{x\to \infty}y(x)$ for the solution with $y(0)=-1$ and the solution with $y(0)=1$. For both of these I simply look at the directional field and follow the lines. So for $y(0)=-1$, I guess $\lim\limits_{x\to \infty}y(x)=-\infty$ and for $y(0)=1$, I guess $\lim\limits_{x\to \infty}y(x)=3$.

  • for what initial conditions the solution $y(x)$ is bounded for all $x\in\mathbb{R}$. Once again I look at the directional field and guess that all solutions with initial conditions under or on the parabola that we see aren't bounded. So I say that all solutions with $y(0)>0$ are bounded.

The way I have "solved" these problems does not feel very nice and I am not sure it is right. I would like to hear if there are ways I could find the things I am asked using the differential equation, without relying as much on the directional field. I would also appreciate suggestions on how to find the solution which corresponds to the parabola (I'm not sure of course it's a parabola, but I think so) that we see.

Directional field corresponding to given differential equation

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In regard to parabola, it has formula $y=\frac{-x^2}{3}$ – unknown Dec 9 '12 at 18:56
@unknown Ok thanks, I think I see that now as well. Is there any chance of you making an answer (assuming you have suggestions for the other things as well)? – user50407 Dec 9 '12 at 19:30
up vote 1 down vote accepted

The second part can be proved rigorously.

  1. If $y_0>0$, then the solution is bounded. Indeed, we have $y'>0$ when $y>0$, which implies that $y$ is an increasing function that stays positive at all times. In the opposite direction, $y'\le \frac{1}{3y_0+x^2}+\frac{2}{4+x^2}$, and since the integral of RHS converges, $y$ stays bounded. More precisely, $$y(x)-y_0\le \int_0^\infty \frac{dx}{3y_0+x^2}+\int_0^\infty \frac{2\,dx}{4+x^2}$$ which you can integrate to get an upper bound on $\lim_{x\to\infty} y(x)$. I'm not very happy that this upper bound becomes huge when $y_0$ is close to $0$. But this defect can be fixed if we recall that the solution lines do not cross. Therefore, a solution with $y_0<1$ will always stay below the solution with $y_0=1$, and therefore will obey the upper bound for $y_0=1$.

  2. If $y_0<0$, we want to prove that the solution stays below the parabola $y=-x^2/3$. One way to do this is to consider the function $u=y+x^2/3$ which satisfies the equation $$u'=\frac{1}{3u}+\frac{2}{4+x^2}+\frac{2x}{3}$$ On any interval $[0,M]$ the RHS is bounded above by $\frac{1}{3u}+B$ where $B$ is a constant that is greater than the sum of two other terms. Let's also make $B$ large enough so that $u(0)<-1/(3B)$. It follows that $u<-1/(3B)$ on the entire interval $[0,M]$, because $u$ cannot go up in the territory where the direction field has negative slope.

The above sounds informal, but can be made rigorous with an appropriate comparison theorem for ODE: e.g., if two functions $u_1,u_2$ have the same initial value at $x=0$ and satisfy $u_1'=f(u_1,x)$ and $u_2'\ge f(u_2,x)$, then $u_2(x)\ge u_1(x)$ for all $x\ge 0$. Here $u_1$ is a solution of ODE while $u_2$ could be called its supersolution (it outperforms the target rate of increase). The concept of subsolution is defined similarly. Sadly, they are not normally found in ODE textbooks which focus on solving equations precisely.

Speaking of being precise, I have no idea about the precise value of $y(\infty)$ for the initial value $y_0=1$.

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Thanks, I think this is just what I wanted. I will take a good look at it. – user50407 Dec 26 '12 at 22:09

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