Question

In a previous exercise we formulated a model for learning in the form of the differential equation dP dt = k(M − P) 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. Solve this differential equation to find an expression for P(t). (Use P for P(t). Assume that P(0) = 0.)

Answer 1

Answer:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

Step-by-step explanation:

For this case we aasume the following differential equation:

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

Is a separable differential equation so we can do the following procedure:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$