Can You Solve This Differential Equation Involving Exponential Decay?

[zheng89120]

Homework Statement

Assume N_A = N_Ao exp(-t/a)

Solve the differential equation:

dN_B/dt + N_B/b = N_A/a

Homework Equations

differential equations

The Attempt at a Solution

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

with initial condition: C + D = N_Bo

I tried plugging in this and N_A into the original equation, but was not able to solve for C or D...

 

[vela]

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.

 

[zheng89120]

Unfortunately I don't really have much experience with this kind of differential equation. After I substituted the trial function and C = N_B - D into the equation, I got:

[ -(N_B-D)/a*exp(-t/a) - D/b*exp(-t/b) ] + [ N_B-D/b*exp(-t/a) - D/b*exp(-t/b) ] = N_A/a*exp(-t/a)

how would you solve 'D' from this?

 

[vela]

You want to plug the trial function into the differential equation:
\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}
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.

 

[zheng89120]

Ok, thanks for the insightful help, vela. This is what I got for N_B, with D=N_Bo-C :

N_B = [N_Ao / (a/b-1)] *exp(-t/a) + [N_Bo - N_Ao / (a/b - 1)] *exp(-t/b)

assuming this is correct.