Can anyone suggest to me a method to solve the following?

[JonoSmith]

I have the following DE that I am trying to integrate. I have been looking at Integrating Factors, but am not sure if it is applicable to this example.

∫〖1/S ds〗= -α∫ PA.e^αPt/ (1+A.e^αPt)

As I said, some tips on a method to do this would be better than solving it for me, because i want to look into it myself and try and see what is going on

Thank you

 

[JonoSmith]

Anyone have any idea?

 

[g_edgar]

Not even an idea what it means.

 

[jackmell]

Hi Jono. Let me be frank with you. That's indecipherable for me and perhaps others too. For one thing you use little s and big S in the same expression on the left. Also, there are some spurious block-characters in your notation which makes it also cryptic. Then you have no differential on the right side in the integral expression. Now I'm new here and I'm not one to criticize even if I'm old but that is why I think no one is responding. So if you could remember to write your math as utterly nice as humanly possible, I think you would get more replies, like for example:

\int \frac{1}{s}ds=-\alpha\int \frac{P a e^{\alpha P t}}{1+A e^{\alpha P t}}dt

if that indeed is what it is suppose to be.

 

[D H]

I'm using it to model an Epidemic

∫〖1/S ds〗= -α∫ PA.e^αPt/ (1+A.e^αPt)

What do your S and s represent?

 

[jackmell]

Cheers Jack, I am also new here and was not aware of this Latex Equation input, will give it a go now to see.

\int \frac{1}{S} dS=-\alpha\int \frac{P A e^{\alpha P t}}{1+A e^{\alpha P t}}dt

Is how it is supposed to look. So from here if anyone is able to give me some help it would be much appreciated.

Very good then. Integrate both sides. Note that if I let u=1+Ae^{\alpha P t} then du=A\alpha P e^{\alpha Pt}dt. Then:

\ln|S|=-\int\frac{1}{u}du

or:

S(t)=K\frac{1}{1+Ae^{\alpha P t}}

where K is a constant of integration dependent on the initial conditions or you could have integrated explicitly:

\int_{S_0}^{S(t)} \frac{1}{S} dS=-\alpha\int_{t_0}^t \frac{P A e^{\alpha P t}}{1+A e^{\alpha P t}}dt

 

[JonoSmith]

Thank you, that makes sense to me.

Cheers so much for the help.