# Help interpreting the solution for a differential equation

The differential equation is

$\frac{dx}{dt} = x + x^2$

Solving for $x$, I got

$x = (ce^t)/(1- ce^t)$

where, $c = x_0/(1+x_0)$ and $x_0$ is the initial value of $x$ at $t=0$

Now, the value of $dx/dt$ is positive for positive $x$ values. Thus, if $x_0$ is greater than zero, then the value of $x$ is expected to monotonously increase.

However, when I look at the solution as time tends to infinity, irrespective of the constant $c$ and $x_0$, the value of $x$ reaches $-1$.

It would be really helpful if you could explain where I am going wrong.

• I think you might want to say, "the value of $x$ is expected to monotonicly increase;" monotonicity is not the same thing as monotony, though it may seem sometimes that monotonic functions behave in a monotonous manner. 😉 Cheers! – Robert Lewis Aug 28 '15 at 2:51

I think the source of confusion is that $\frac{dx}{dt}$ is given in terms of $x$, not $t$.
Thus, it is true that, as $x$ gets bigger, the slope also gets bigger. However, if $x_0$ is positive, then the slope gets larger and larger until we reach the point that $ce^t=1$, when the slope is undefined. After that, the value of the function is negative, and does in fact tend towards $-1$ as $t$ goes to infinity.
Think about it this way: you can rewrite the equation as $\frac{dx}{dt}=x(1+x)$ then graph this polynomial $f(x)=x(1+x)$. Where $f(x)$ is positive, $x$ is an increasing function of $t$. Thus is can be concluded that $x$ is an increasing function of $t$ when $x<-1$ or $x>0$, and that $x$ is a decreasing function of $t$ for $-1<x<0$.
So if you have a positive initial value $x(0)>0$, then as $t$ tends to infinity, $x$ will grow unboundedly. If your initial value is in the interval $-1<x(0)<0$, then $x(t)$ tends to $-1$ as $t$ increases. The same thing happens is $x(0)<-1$.