Trouble with Solving a Differential Equation

In summary: As t goes to infinity, e^{-kt} goes to 0 so the limit is M- 0= M.In summary, we have a model for learning in the form of a differential equation, where P(t) measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. The basic approach for solving the differential equation is to "separate and integrate". The function P(t) can be found by dividing by M-P and integrating both sides, and the final solution is P(t) = M - Me^{-kt}. The limit of this expression as t goes to infinity is M, regardless of the constant C.
  • #1
xxdrossxx
8
0
I'm having trouble with the following textbook problem:

A model for learning in the form of a differential equation is seen below:
[tex]\frac{dP}{dt} = k(M - P)[/tex]
where [tex]P(t)[/tex] measures the performance of someone learning a skill after a training time [tex]t[/tex], [tex]M[/tex] is the maximum level of performance, and [tex]k[/tex] is a positive constant. Solve this differential equation for [tex]P(t)[/tex]. What is the limit of this expression?

The only basic approach for solving differential equations that I know is to "separate and integrate", that is, have one side with P's and a dP, and the other side with t's and a dt (constants anywhere throughout), and then integrate both sides. So, what I was hoping to get was kM * something * dt = kP * something * dP. However, it looks like that's not going to happen. Could someone help me solve this thing?

I know that if "M is the maximum level of performance", then logically [tex]\lim_{t \rightarrow \infty} P(t) = M[/tex]. However, I'm still not sure how to actually find the function P(t).

Thanks for any help you can give. :smile:
 
Last edited:
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  • #2
xxdrossxx said:
I'm having trouble with the following textbook problem:



The only basic approach for solving differential equations that I know is to "separate and integrate", that is, have one side with P's and a dP, and the other side with t's and a dt (constants anywhere throughout), and then integrate both sides. So, what I was hoping to get was kM * something * dt = kP * something * dP. However, it looks like that's not going to happen. Could someone help me solve this thing?

I know that if "M is the maximum level of performance", then logically [tex]\lim_{t \rightarrow \infty} P(t) = M[/tex]. However, I'm still not sure how to actually find the function P(t).

Thanks for any help you can give. :smile:


It's easily separable just divide by by M-P and you have a pretty standard integral to find the function P.
 
  • #3
d_leet said:
It's easily separable just divide by by M-P and you have a pretty standard integral to find the function P.
Ah, of course! I guess my first mistake was distributing the k. Dividing brings me to [tex]\frac{dP}{M - P} = kdt[/tex], and integrating both sides gives me [tex]-ln|M - P| = kt + c[/tex]. Going from there:

[tex]ln|M - P| = -kt - c[/tex]
[tex]e^{-kt - c} = M - P[/tex]
[tex]P = M - e^{-kt - c}[/tex]
I think I'm doing something very wrong here. The answer in the back of the book is [tex]P(t) = M - Me^{-kt}[/tex].

I'm thinking I've made a really stupid mistake, but I can't find it. Can anyone help?
 
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  • #4
Have you left out part of the problem- like, say, the information that the subject knew nothing to start with? That is, that P(0)= 0.

[tex]P(t) = M - e^{-kt - c}= M- (e^{-c})e^{-kt}[/tex]
and we can write that as
[tex]P(t)= M- C e^{-kt}[/tex]
where [itex] C= e^{-c}[/itex].

Then, if P(0)= 0,
[tex]P(0)= M- Ce^0= M- C= 0[/tex]
so C= M.

Actually that's not relevant to the final question: what is the limit of P(t) as t goes to infinity? That answer does not depend on C.
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It involves one or more variables and their rates of change, and is commonly used in physics, engineering, and other fields to model real-world phenomena.

Why is it difficult to solve a differential equation?

Solving a differential equation can be challenging because it often involves complex mathematical operations and techniques. In many cases, there is no analytical solution and numerical methods must be used. Additionally, the behavior of the solution may be sensitive to small changes in the initial conditions or parameters, making it difficult to obtain an accurate solution.

What are some common techniques for solving differential equations?

Some common techniques for solving differential equations include separation of variables, substitution, and integration by parts. Other methods such as power series, Laplace transforms, and numerical methods like Euler's method can also be used depending on the type of differential equation.

What are initial conditions in a differential equation?

Initial conditions refer to the values of the dependent and independent variables at a specific starting point, usually denoted as t = 0. These conditions are necessary for solving a differential equation as they provide the starting point for finding the solution.

How are differential equations used in scientific research?

Differential equations are used in a wide range of scientific research to model and study various phenomena, such as chemical reactions, population growth, and fluid dynamics. They provide a powerful tool for making predictions and understanding the behavior of complex systems in the natural world.

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