1
$\begingroup$

I'm solving a question from my Calculus textbook that asks the following:

Let $P(t)$ be the performance level of someone learning a skill as a function of the training time $t.$ The graph of $P$ is called a learning curve. In Exercise 9.1.15 we proposed the differential equation

$\frac{dP}{dt} = k[M-P(t)]$

as a reasonable model for learning, where $k$ is a positive constant. Solve it as a linear differential equation and use your solution to graph the learning curve.

My solution was to use the fact that a linear differential equation can be written in the following form:

$\frac{dy}{dx} + yP(x) = Q(x)$

where $t$ is the independent variable, and $P$ is the dependent variable (so $P(t)$ is the like 'y' in this case.)

I rewrote the given differential equation for the performance level as such:

$\frac{dP}{dt} + kP(t) = kM$

Thus, my $k$ is like my "function of x", and my $P(t)$ is my "y."

By setting $e$ to the power of the integral of $k$, I got $e^{kt}$ as my integration factor.

I multiplied both sides of my differential equation by $e^{kt}$, and then rewrote the left side as $(e^{kt} · P(t))'$. I integrated both sides, and factored out $k$ and $M$ since they're both constants.

The answer I got for $P(t)$ is $M + ce^{-kt}.$

So for my actual question: Is this correct? I'm not very sure about my solution to this question. Also, is my logic for each of my steps sound and reasonable?

Thank you so much for reading!

$\endgroup$
  • $\begingroup$ Your answer solves the differential equation, as you can check by differentiation. However, I suppose we are to have $P(0)=0,$ and you haven't reflected that. What should the value of $c$ be? $\endgroup$ – saulspatz Mar 30 '18 at 23:33
  • $\begingroup$ I think that if we are assuming $P(0) = 0,$ then $c$ should probably be $-M.$ $\endgroup$ – DeepLearner Mar 31 '18 at 1:12
  • $\begingroup$ That's what I think, too. $\endgroup$ – saulspatz Mar 31 '18 at 1:55
0
$\begingroup$

What you did is correct. The equation can also be solved as a separable equation: you can write $$ \frac{P'}{M-P}=k. $$ Looking at antiderivatives you get $$ -\log(M-P)=kt+c, $$ and taking exponentials $$ M-P=e^{-kt-c}=de^{-kt}. $$ So $$ P(t)=M-de^{-kt}. $$ Since the constant $d$ is still to be determined, the minus sign can be made into a plus to get the solution in the same form as you had.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.