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.) incorrect: your answer is incorrect.

Answer 1

I assume you mean

   $\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

   $P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

   $\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt \end{aligned}$

Integrate both sides of the equation.

   $\begin{aligned} \int \dfrac{dP}{M-P} &= \int k\, dt \\ -\ln|M-P| &= kt + C \\ \ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us 

   $u = M - P \implies du = -1dP \implies dP = -du$

hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is 

   $\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

   $\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ M - P &= \pm e^{-kt - C} \end{aligned}$

Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

   $\begin{aligned} M - P &= \pm e^{-kt - C} \\ M - P &= \pm e^{- C - kt} \\ M - P &= \pm e^{- C + (- kt)} \\ M - P &= \pm e^{- C} \cdot e^{- kt} \\ M - P &= Ke^{- kt} && (\text{\footnotesize Let K = \pm e^{-kt} }) \\ M &= Ke^{- kt} + P\\ P &= M - Ke^{- kt} \end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

   $\begin{aligned} 0 &= M - Ke^{- k\cdot 0} \\ 0 &= M - K \cdot 1 \\ M &= K \end{aligned}$

Substituting into our original equation, we get our final answer of

   $P = M - Me^{-kt}$

Answer 2

Hi there!

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

⇒ $\dfrac{dP}{M-P} = k$

$\textbf{Integrate\: both \:sides\: of\: th'\: Eqn.}$ :-

$\int \dfrac{dP}{M-P} = \int kdt$

⇒ $\ln|M-P| = kt + C$

⇒ $\ln|M-P| = -kt - C$

Since,
$u = M - P \implies du = -1dP \implies dP = -du$

∴ $\int \frac{dP}{M-P} \ne \ln|M - P|$

$\textbf{Applying\: definition \:of\: logarithm}$ :-

$\ln|M-P| = -kt - C$

⇒ $|M - P| = e^{-kt - C}$

⇒ $M - P = \pm e^{-kt - C}$

$\textbf{Further solving}$ :-

⇒ $M - P = \pm e^{-kt - C}$

⇒ $M - P = \pm e^{- C - kt}$

⇒ $M - P = \pm e^{- C + (- kt)}$

⇒$M - P = \pm e^{- C} × e^{- kt}$

⇒ $M - P = Ke^{- kt}$

⇒ $M = Ke^{- kt} + P$

⇒ $P = M - Ke^{- kt}$

$\textbf{Let \:P(0) \:= \:0, \:then \:set\: t\: =\: 0\: and\: P \:= \:0}$ :-

$0 = M - Ke^{- k × 0}$

⇒ $0 = M - K × 1$

⇒ $M = K$

Substitute th' following value in Original Eq :-

$\boxed{P = M - Me^{-kt}}$


~ Hope it helps!