solution to an equation (1 Viewer)

[Loner]

Can it be shown that x = a - Ae^kt satisfies dx/dt = k(a - x)? I trued this by substution and it doesn't work, and it's supposed to. When I did it, I found k should be negative in the first equation to make it work. Can anyone help out here?

Thanks..

 

[watatank]

x = a - Ae^kt

differentiate first equation: dx/dt = kAe^kt

rearrange first equation: Ae^kt = (a - x)

use those two things to show

dx/dt = k(a - x)

 

[Loner]

x = a - Ae^kt

differentiate first equation: dx/dt = kAe^kt

rearrange first equation: Ae^kt = (a - x)

use those two things to show

dx/dt = k(a - x)

shouldn't dx/dt be -kAe^kt ? (what happened to the minus sign in front of the A?

 

[watatank]

hmm youre right about that. maybe its only valid for certain values of k. this is me just guessing here btw.

 

[Loner]

haha thanks. I think the question given to me is wrong. Substitution and checking is the first part of the question. In the second part, I found k is negative and subsequently found an answer for the question. Anyone else got any ideas?

 

[Loner]

Noone else? Justy wanna know if x = a - Ae^kt satisfies dx/dt = k(a - x). Both with integration and the substitution method i found it doesn't satisfy the equation. If it does please show me how! *lol*

 

[pLuvia]

x=a-Ae^kt
dx/dt=-kAe^kt
Since Ae^kt=a-x
then dx/dt=-k(a-x)

If k is negative then dx/dt=k(a-x)
If k is positive then dx/dt=-k(a-x)

 

[Loner]

x=a-Ae^kt
dx/dt=-kAe^kt
Since Ae^kt=a-x
then dx/dt=-k(a-x)

If k is negative then dx/dt=k(a-x)
If k is positive then dx/dt=-k(a-x)

Ok thanks for that it makes sense now. K should be negative in my question,