Yes, that is a linear equation but it is also "separable" and I think that is an easier approach.
Separate the variables:
and each side can be integrated (on the left let ).
I'm trying to solve analytically the following simple first order differential equation:
for and being constants. In principle, this differential equation is linear, right? However, the only thing I could think at the moment is to try to write it differently to approximate to the following expression:
that has a general solution of the form:
For that, what I did so far:
1. Reorganize the formula by multiplying everything by :
2. Group the terms involving the unknown:
3. Reorganizing we can write as:
4. Dividing by
This gives me , and which are constants, and not functions.
Is this the right way to proceed or should I try another method?
Thanks for your help
Wov! Thanks!
I first tried to do it with by separation of variables, but I wasn't able to separate the equation into 2 expressions:
From your separated expression:
Let be ,
then ,
both expressions that we substitute in the differential equation:
Reorganize the negative symbol:
and moving the constants to the left side:
then solving for y:
and applying exponentiation to both sides...
I'm not very sure this is the right answer....
Hey Melody2!!!
Unless I'm doing something wrong, your solution
is not correct.
I've just tested whether the solution of that my CAS provide give the same return values as your equation with the same parameters (K=1, U=0.8 and tau=2) for some values(for example for t=1 and t=5), and they're different. Strange enough, the steady-state conditions match.
See details here
my solution above doesn't work either.
Hi
Actually, I'm not very sure about that, for 2 reasons:
1) Your equation differs from the one that results from the symbolic integration computed with software (see here)
2) More importantly, note that the solution of the differential equation in steady-state depends on and , since
,
then ,
operating and organizing we have
,
,
,
solving for x:
,
when your equation tends to infinity, the value of x tends to zero, because
Now I know that the solution is like this:
But do not know how to get there... or to get a more simple expression
I think it will help me to try to solve it again:
From my (correct?) expression:
We apply integration to the previous expression:
and solve that integral, we obtain:
being the integration constant. Now we have to apply exponential to both sides. In this the step I'm not very sure if the exponential of the left hand side is correct (I assumed that exponential of a sum is not the same as the sum of the exponentials).
Since we know that we can write:
reorganizing everything:
which results in this monster:
I missed something...
Hi jguzman,
My knowledge has holes in it and I actually didn't understand what you meant by steady-state. I can see that as t gets bigger f(t) appears to be or approach the same constant. Maybe 0.4 I assume that that is what you meant.
I am also interested in the Sage code that you used. Is that program an internet freeby?
Melody.
Hi Melody!
That's exactly what I meant by steady-state. The differential equation has an steady-state if the function x(t) does not change anymore with the independent variable (in my case t). In that case, the rate of change of x(t) is zero, therefore the differential equation is zero. The value that I obtained in the steady state dependended on both and . The relationship is:
For the numerical computations I performed, I set and , so we have an steady-state of 0.38
Both yours and mysolution have the same steady state, so I guess they only differ in the way the constants are defined. The definition of the constants is important in my case because I'm trying to model a biophysical system.
Yes, Sage is free. In principle you can use it in your computer, via a cloud or in Internet via the link I sent you (this allows you to share pieces of code with colleagues, as I did here).
I will send you a link via a private message and some examples, ok?