# Linear Differential Equation for Performance Level

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!

• 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? – saulspatz Mar 30 '18 at 23:33
• I think that if we are assuming $P(0) = 0,$ then $c$ should probably be $-M.$ – DeepLearner Mar 31 '18 at 1:12
• That's what I think, too. – saulspatz Mar 31 '18 at 1:55

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.