Incorrect differential equation problem

Question

I'm going through James Stewart Calculus 8th edition and I cannot figure out what I'm doing wrong on this problem.

I am asked to solve to find an equation for P(t) given

(dP/dt) = k(M - P)

P(t) models performance of someone learning a skill after time t. M is the max performance and k is a constant.

(M - P)^-1 * dP = k * dt

∫ (M - P)^-1 dP = k∫ dt

-ln(M - P) = kt M - P = e^-kt

P(t) = M - e^-kt

However, the back of the book lists:
P(t) = M - Me^-kt
as the answer.

I have no idea where the second M comes from. It also seems they didn't have a constant (as in C after performing an integral). Is this because of k? Or because they assume P(0) = 0 ? Why am I not getting their answer?

Any help would be greatly appreciated! Thanks!

Answer 1

You have$$\int(M-P)^{-1}\,dP=k\int dt$$ This gives $$M-P=e^{-k(t-t_0)}$$ where $t_0$ is a constant of integration. Then we get$$ P=M-e^{-k(t-t_0)}=M-e^{-kt}e^{kt_0}=M-\alpha e^{-kt}$$where we define $\alpha=e^{kt_0}$. If you assume $P(0)=0$, then you get $$P(0)=0=M-\alpha\\\implies \alpha=M$$ Which would give $$P=M-Me^{-kt}$$

Answer 2

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

and integrating

$$ \ln|M-P| = -k t - C_0\Rightarrow P = M-C_1e^{-kt} $$

now if $P(0) = 0$ then $C_1 = M$

and finally

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