Solving DE x'(t) = -.20x(t) to Find Amount After 1 Year

[Sheneron]

Homework Statement

Say something has an amount and it decreases by 20% a year. Let's say the starting amount is 100. Then, after 1 year youd expect the amount to be 80.

The DE x'(t) = -.20x(t) would represent the above comment. I think. (x(t) is a function of the amount after time t.)

However, if you solved that DE and found your function x(t) and plugged in 1 for t, you would not get 80.

Why is this so?

 

[jacobrhcp]

hm... this question makes me feel really stupid since I don't know, and I'm 3rd year math&physics major.

However, if you assume your DE solution is of the form x'(t)=-a x(t) and get solution x(t)=100e^(-a t)... then you can just solve this for a such that it fits x(1)=80.

I know it's a stupid solution, but the best I can offer you... and since I just tried it in mathematica I know it will fit your expectations for x(2) and x(3) also. I somehow managed never to look at diffeq. in this realistic of a way...

 

[Sheneron]

Thanks for the reply. If I did as you said though, and solved x(t) = 100e^(-at) so that x(1) = 80, then it wouldn't decrease at 20% as stated previously. So it can either decrease at 20% or equal 80, but not both. Which is odd because logically it would 80. The DE isn't wrong is it? I would be curious to know the explanation as well.

 

[HallsofIvy]

Homework Statement

Say something has an amount and it decreases by 20% a year. Let's say the starting amount is 100. Then, after 1 year youd expect the amount to be 80.

The DE x'(t) = -.20x(t) would represent the above comment. I think. (x(t) is a function of the amount after time t.)

However, if you solved that DE and found your function x(t) and plugged in 1 for t, you would not get 80.

Why is this so?

"decreases by 20% a year" means that x(n+1)= x(n)- 0.2 x(n)= 0.8x(n), a difference equation. The differential equation you give, assumingntinuous decrease corresponds to an annual decrease of slightly more: about 22.3%. It is exactly the same as noting that if you put money in an account at 20% compounded continuously it will earn the same interest as an acount giveing simple interest of 22.3%

 

[Sheneron]

Thank ye.

 

[Sheneron]

So now what if I have a question that says: Something loses a certain percent of value per year. Write a DE and solve. If something loses a certain percent value per year, then the rate of decrease wouldn't be proportional to itself and there wouldn't be a DE would there?

 

[HallsofIvy]

No, it is "proportional" but with a different proportionality.

Suppose you know that quantity x loses "k" percent per year. That is the same as saying that dx/dt= -ax but a is NOT equal to "k/100". The solution to dx/dt= -ax, with x(0)= 1 is, as you say, x(t)= e^-at. After 1 year that will be x(1)= e^-a so that the amount lost was 1- e^-a= k/100 That is the same as e^-a= 1- k/100 so -a= ln(1- k/100)= ln((100-k)/100) or a= ln(100)- ln(100-k).

Notice that after the second year, x(2)= e^-2a so the x has decreased from e^a to e^-2a= e^-a- (e^-a)[sup²= x1- x1( e^-a). That is, during the second year x had decreased by the same proportion of its vaue as it did the first year.

 

[Sheneron]

For an example: You have a computer that loses 20% of its value per year.

Would this equation not work? x(t+1) = x(t)(0.8)^t. Thats not a DE but that would work would it not?

Also, let me see if I get what your saying. The DE dx/dt = -ax (where a is a constant, but not equal to -0.2), and you have to solve for a, based on conditions?

 

[HallsofIvy]

For an example: You have a computer that loses 20% of its value per year.

Would this equation not work? x(t+1) = x(t)(0.8)^t. Thats not a DE but that would work would it not?

If you mean x(t+1)= (0.8)x(t), without the "^t", yes, that would be a "difference equation" that would give the correct value for t integer valued years.

Also, let me see if I get what your saying. The DE dx/dt = -ax (where a is a constant, but not equal to -0.2), and you have to solve for a, based on conditions?[/QUOTE]
Since you were asking about the percentage change, you can take any a(0) you like. The general solution to it is x(t)= x(0)e^-at. Then x(1)= x(0)e^-a so that x(1)/x(0)= e^-a. In fact, for any two succesive years, x(t)= x(0)e^-at and x(t+1)= x(0)e^-at-a and dividing one by the other x(t+1)/x(t)= e^-a. IF a quantity loses 20% of its value per year, then that proportion would be 0.80 so you must have e^-a= .8 or a= -ln(.8) which is approximately .223. In general, if a quantity loses "k" percent of its value each year, then the proportion is 1- k/100= (100- k)/100 so we have e^-a= (100- k)/100 as I just said. Given k, you can solve that for a.

 

[Sheneron]

So the moral of the story is that if you say something loses a certain percent per year, then in the DE the proportionality constant is not quite that value. You have to solve for "a"

to clarify:

if something loses 20% of its value each year then the DE would not equal x'(t) = -.20x(t)