Diferential Equation by Laplace: $y'=-\frac{y}{ay+b}+S$

Question

Hello and thanks for reading.

I wish to check the stability of the ODE

$$ y'=-\frac{y}{ay+b}+S $$

by applying the limit theorem of Laplace, where $a$, $b$ and $S$ are real. After some algebra I have:

$$ ay'y+by'-aSy-bS+y=0. $$

When I do Laplace transformation then I have to $L(y'y)$ and can't go on. I need the Laplace expression of $y$.

Thanks for your time.

Answer 1

The asymptotic behavior of $y(t)$ for $t \to \infty$ can be qualitatively modeled without the need of an explicit solution of the ODE:

$$ \frac{{\rm d}y}{{\rm d}t} = -\frac{y}{ay + b} + S\tag{1} $$

First of all, consider two cases

CASE 1: $\color{red}{a=0}$

In this situation (1) becomes

$$ y(t) = -\frac{y}{b} +S $$

whose solution is

$$ y(t)= b S+(y(0) - b) e^{-t/b} $$

So you have two cases

$a = 0$, $b > 0$

$$ \lim_{t\to \infty}y(t) = bS $$

regardless of the initial condition $y(0)$

$a = 0$, $b < 0$

$$ \lim_{t\to \infty}y(t) = +\infty $$

regardless of the initial condition $y(0)$.

The case $a=0$, $b=0$ I will not consider because it leads to problems from the beginning.

CASE 2: $\color{red}{a\not=0}$

If this is case I can call $t' = t/a$, $b' = b/a$ and $S' = Sa$ so that the (1) becomes

$$ \frac{{\rm d}y}{{\rm d}t'} = -\frac{y}{y + b'} + S' \tag{2} $$

Intuitively speaking, if the RHS of this equation is 0, then the solution does not evolve from there. If it is larger than zero the solution grows, and decreases with time otherwise. So let us consider the situation

$$ -\frac{y^*}{y^* + b'} + S' = 0 ~~~\Rightarrow y^* = \frac{S'b'}{1 - S'} = \frac{Sb}{1 - aS} \tag{3} $$

If $y(0) = y^*$ it means that $y(t) = y(0)$ and

$$ \lim_{t\to \infty}y(t) = \frac{Sb}{1 - aS} $$

In general we can expand $f(y)$ around the point $y^*$

$$ y = f(y^*) + \left.\frac{{\rm d}f}{\rm d y}\right|_{y = y^*}(y - y^*) + \cdots $$

If we cal $\zeta = y - y^*$, at first order this equation implies

$$ \frac{{\rm d}\zeta}{\rm d t} = \alpha \zeta(t) ~~~\mbox{with}~~~ \alpha = \left.\frac{{\rm d}f}{\rm d y}\right|_{y = y^*} = \frac{(a S - 1)^2}{b} $$

There are a few cases to consider here

$S \not= 1 /a$, $b > 0$

For this case $d\zeta/dt > 0$, if $\zeta > 0$, which means that the separation between $y$ and $y^*$ grows exponentially.

The rest of the analysis I will leave for you to finish