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[SOLVED] Possible to integrate a function f(x) which is equal to the function itself?

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mandom

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Hi all,

I got stuck half way when trying to solve/prove an equation as listed in the attachment.

Here, I need to find the rate of change of current function Ip with respect to z direction, which is equal to
gBIp2 + gBIoIpexp(-az) - aIp

whereby gB, Io and a are constants.

For what I understand so far, the author is trying to make the equation integrable by dividing both side with -Ip2 , which then leads to second equation in the attachment.

Upon direct integration w.r.t z, I'll get 1/Ip on the LHS (last equation). But I'm puzzled with how to reach RHS.

1. First, I notice that on the LHS I already have 1/Ip, on the RHS I need to integrate 1/Ip again, which got me thinking if this is some sort of recursive?

2. I've tried integration by parts but how can I possibly eliminate Ip totally in this case?

Thanks for any hints in advance.

Regards,
Kok Kuen
 

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I'm a bit confused. The first equation IS the rate of change of IP wrt z. Isn't that what you said you wanted to solve? But, if you feel the need to integrate it you can do that, and then solve for Ip using the quadratic formula.
 

Sorry for the confusion. I should have mentioned I need to solve for Ip given the rate of change of IP wrt z equation. How does applying quadratic formula work here?
 

I think you can proceed as follows. To simplify the writing let's call:

y=Ip
x=z
A=gb
B=gbIo
C=a

then you equation is:

dy/dx=A•y²+B•y•exp(-α•x)+C•y

rearranging:

dy/dx-A•y²-[B•exp(-α•x)+C]•y=0

This is a Riccati's differential equation that can be solved. Try googling "riccati's equation" or the very useful site https://www.wolframalpha.com/
 
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    mandom

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I think you can proceed as follows. To simplify the writing let's call:

y=Ip
x=z
A=gb
B=gbIo
C=a

then you equation is:

dy/dx=A•y²+B•y•exp(-α•x)+C•y

rearranging:

dy/dx-A•y²-[B•exp(-α•x)+C]•y=0

This is a Riccati's differential equation that can be solved. Try googling "riccati's equation" or the very useful site https://www.wolframalpha.com/

Hi albbg,

You are the man. I can't thank you enough for this.
 

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