Can You Solve This Differential Equation Involving Exponential Decay?

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Homework Help Overview

The discussion revolves around solving a differential equation related to exponential decay, specifically involving the equation dNB/dt + NB/b = NA/a, where NA is defined as NA = NAo exp(-t/a). Participants are exploring methods to find a solution for NB.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of a trial function for NB and the implications of initial conditions. There are attempts to substitute this trial function into the differential equation, with some participants questioning how to isolate constants C and D from the resulting expressions.

Discussion Status

Guidance has been provided regarding the cancellation of terms when substituting the trial function into the differential equation. Some participants are exploring the implications of these cancellations and how they affect the ability to solve for constants. Multiple interpretations of the problem setup are being discussed, with no explicit consensus reached yet.

Contextual Notes

One participant notes a lack of experience with this type of differential equation, which may influence their approach and understanding. The discussion includes assumptions about the behavior of the terms in the equation and their contributions to the solution.

zheng89120
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Homework Statement



Assume NA = NAo exp(-t/a)

Solve the differential equation:

dNB/dt + NB/b = NA/a

Homework Equations



differential equations

The Attempt at a Solution



trial function: NB = C exp(-t/a) + D exp (-t/b)

with initial condition: C + D = NBo

I tried plugging in this and NA into the original equation, but was not able to solve for C or D...
 
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Show us what you got because what you described should work.

Note that the D term will vanish when you plug it in because it's the solution to the homogeneous differential equation. This allows you to solve for C.
 
Unfortunately I don't really have much experience with this kind of differential equation. After I substituted the trial function and C = NB - D into the equation, I got:

[ -(NB-D)/a*exp(-t/a) - D/b*exp(-t/b) ] + [ NB-D/b*exp(-t/a) - D/b*exp(-t/b) ] = NA/a*exp(-t/a)

how would you solve 'D' from this?
 
You want to plug the trial function into the differential equation:
[tex]\frac{d}{dt}(C e^{-t/a} + D e^{-t/b}) +\frac{1}{b}(C e^{-t/a} + D e^{-t/b}) = \frac{N_{A_0}}{a} e^{-t/a}[/tex]
When you differentiate the first term, you'll see the e-t/b terms cancel out, which leaves only the e-t/a terms, allowing you to solve for C. Once you know C, you can solve for D.
 
Ok, thanks for the insightful help, vela. This is what I got for NB, with D=NBo-C :

NB = [NAo / (a/b-1)] *exp(-t/a) + [NBo - NAo / (a/b - 1)] *exp(-t/b)

assuming this is correct.
 

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