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when do we have to look for a constant solution to the differential equation?

Example: Solve the initial value problem $\frac{dy}{dx} = \frac{2}{\sin y}$ with the condition $y(0)=0$.

The first step in the worked solution is as follows: Noting that $\frac{1}{\sin y} ≠ 0$ for any $y$, we see there are no constant solutions.

My question is: do we always have to look for constant solutions when solving any differential equations? and what is the significance of the constant solution? thank you

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The significance of constant solutions - is the fact, they are constant. In order to find them, you need only to find the zeros of the right hand side, i.e. they are - in general case - easy to find.

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Constant solutions/the zeros of the right hand side of the differential equation are also very useful in understanding the QUALITATIVE behavior of more general solutions to a differential equation. That is why I always look for constant solutions first if I want to understand an ordinary differential equation. Consider $$ y'(t) = y(t) - y^2(t) $$ as an example, where one should think of $t$ as time. The right hand side is analytic, so the standard local existence- and uniqueness theorems apply. Therefore, we know beforehand, that a smooth solution exists for any finite initial value y(0) for SMALL values of our time $t$.

We do not know for example, that there is a smooth solution for any initial value and an arbitrary long time $t$ and in fact this is wrong.

Here, our constant solutions help: Our differential equation has two of them, $y(t) \ = \ 0$ and $y(t) \ = \ 1$, found by finding the zeros of the right hand side. This implies, that all solutions with initial values in between 0 and 1 STAY THERE, because due to local uniqueness they CANNOT cross the constant solutions. This also implies they exist for arbitrary long times as our right hand side is smooth. Futhermore the right hand side is positive there, so all solutions in this region are INCREASING.

For initial values above 1 or below 0, the right hand side is negative, so all solutions in those regions are DECREASING.

This implies that our constant solution $y(t) \, = \, 1$ is stable in the sense that all solutions starting with $y(0) \, > \, 1$ converge to it with increasing time $t$, while nearby solution with $y(0) \, < \, 1$ also converge to it with increasing $t$. Here, nearby means $y(0) \, > \, 0$, as the other constant solution is, well, constant, so cannot converge to $y(t) \, = \, 1$.

The constant solution $y(t) \, = \, 0$ is unstable in the sense that all solutions starting sufficiently near to it, depart from it with increasing time $t$. If $y(0) \, < \, 0$, the solutions do not even exist for all times $t$, they have a finite BLOW-UP TIME, where they diverge to $-\infty$.

We see, that the non-constant solutions can be seperated into three different classes which are seperated precisely by the constant solutions.

This example is not a strange exception, but an instance of general phenomena, which are investigated in the qualitative theory of differential equations. There is a huge literature on the subject.

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  • $\begingroup$ "due to local uniqueness they CANNOT cross the constant solutions" please elaborate, i didn't get it $\endgroup$ Apr 21, 2019 at 18:43
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    $\begingroup$ @theenigma017 If two trajectories were to cross, there would be a point, so a certain time $t$, where they cross. At this point in time $t$, two trajectories would continue into the future from the same starting point, which contradicts uniqueness. $\endgroup$ Jun 2, 2019 at 9:09
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Constant solutions also are important to identify as special cases for solution techniques like separation of variables. In $$y'=1-y^2$$ the constant solutions $y\equiv \pm 1$ correspond to the zeros to avoid of the denominator in $$ \int\frac{dy}{1-y^2}=\int dx. $$

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