Respuesta :
I assume you mean
[tex]\dfrac{dP}{dt} = k(M-P)[/tex]
ANSWER
An expression for P(t) is
[tex]P = M - Me^{-kt}[/tex]
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
[tex]\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt \end{aligned}[/tex]
Integrate both sides of the equation.
[tex]\begin{aligned} \int \dfrac{dP}{M-P} &= \int k\, dt \\ -\ln|M-P| &= kt + C \\ \ln|M-P| &= -kt - C\end{aligned}[/tex]
Note that for the left-hand side, u-substitution gives us
[tex]u = M - P \implies du = -1dP \implies dP = -du[/tex]
hence why [tex]\int \frac{dP}{M-P} \ne \ln|M - P|[/tex]
Now we use the definition of the logarithm to convert into exponential form.
The definition is
[tex]\ln(a) = b \iff \log_e(a) = b \iff e^b = a[/tex]
so applying it here, we get
[tex]\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ M - P &= \pm e^{-kt - C} \end{aligned}[/tex]
Exponent properties can be used to address the constant C. We use [tex]x^{a} \cdot x^{b} = x^{a+b}[/tex] here:
[tex]\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}[/tex]
If we assume that P(0) = 0, then set t = 0 and P = 0
[tex]\begin{aligned} 0 &= M - Ke^{- k\cdot 0} \\ 0 &= M - K \cdot 1 \\ M &= K \end{aligned} [/tex]
Substituting into our original equation, we get our final answer of
[tex]P = M - Me^{-kt}[/tex]
[tex]\dfrac{dP}{dt} = k(M-P)[/tex]
ANSWER
An expression for P(t) is
[tex]P = M - Me^{-kt}[/tex]
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
[tex]\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt \end{aligned}[/tex]
Integrate both sides of the equation.
[tex]\begin{aligned} \int \dfrac{dP}{M-P} &= \int k\, dt \\ -\ln|M-P| &= kt + C \\ \ln|M-P| &= -kt - C\end{aligned}[/tex]
Note that for the left-hand side, u-substitution gives us
[tex]u = M - P \implies du = -1dP \implies dP = -du[/tex]
hence why [tex]\int \frac{dP}{M-P} \ne \ln|M - P|[/tex]
Now we use the definition of the logarithm to convert into exponential form.
The definition is
[tex]\ln(a) = b \iff \log_e(a) = b \iff e^b = a[/tex]
so applying it here, we get
[tex]\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ M - P &= \pm e^{-kt - C} \end{aligned}[/tex]
Exponent properties can be used to address the constant C. We use [tex]x^{a} \cdot x^{b} = x^{a+b}[/tex] here:
[tex]\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}[/tex]
If we assume that P(0) = 0, then set t = 0 and P = 0
[tex]\begin{aligned} 0 &= M - Ke^{- k\cdot 0} \\ 0 &= M - K \cdot 1 \\ M &= K \end{aligned} [/tex]
Substituting into our original equation, we get our final answer of
[tex]P = M - Me^{-kt}[/tex]
Hi there!
[tex]\dfrac{dP}{dt} = k(M-P)[/tex]
⇒ [tex]\dfrac{dP}{M-P} = k [/tex]
[tex]\textbf{Integrate\: both \:sides\: of\: th'\: Eqn.}[/tex] :-
[tex]\int \dfrac{dP}{M-P} = \int kdt[/tex]
⇒ [tex]\ln|M-P| = kt + C[/tex]
⇒ [tex]\ln|M-P| = -kt - C[/tex]
Since,
[tex]u = M - P \implies du = -1dP \implies dP = -du[/tex]
∴ [tex]\int \frac{dP}{M-P} \ne \ln|M - P|[/tex]
[tex]\textbf{Applying\: definition \:of\: logarithm}[/tex] :-
[tex]\ln|M-P| = -kt - C[/tex]
⇒ [tex] |M - P| = e^{-kt - C}[/tex]
⇒ [tex]M - P = \pm e^{-kt - C} [/tex]
[tex]\textbf{Further solving}[/tex] :-
⇒ [tex]M - P = \pm e^{-kt - C}[/tex]
⇒ [tex]M - P = \pm e^{- C - kt}[/tex]
⇒ [tex]M - P = \pm e^{- C + (- kt)}[/tex]
⇒[tex]M - P = \pm e^{- C} × e^{- kt}[/tex]
⇒ [tex]M - P = Ke^{- kt}[/tex]
⇒ [tex]M = Ke^{- kt} + P[/tex]
⇒ [tex]P = M - Ke^{- kt}[/tex]
[tex]\textbf{Let \:P(0) \:= \:0, \:then \:set\: t\: =\: 0\: and\: P \:= \:0}[/tex] :-
[tex]0 = M - Ke^{- k × 0}[/tex]
⇒ [tex] 0 = M - K × 1[/tex]
⇒ [tex]M = K[/tex]
Substitute th' following value in Original Eq :-
[tex]\boxed{P = M - Me^{-kt}}[/tex]
~ Hope it helps!
[tex]\dfrac{dP}{dt} = k(M-P)[/tex]
⇒ [tex]\dfrac{dP}{M-P} = k [/tex]
[tex]\textbf{Integrate\: both \:sides\: of\: th'\: Eqn.}[/tex] :-
[tex]\int \dfrac{dP}{M-P} = \int kdt[/tex]
⇒ [tex]\ln|M-P| = kt + C[/tex]
⇒ [tex]\ln|M-P| = -kt - C[/tex]
Since,
[tex]u = M - P \implies du = -1dP \implies dP = -du[/tex]
∴ [tex]\int \frac{dP}{M-P} \ne \ln|M - P|[/tex]
[tex]\textbf{Applying\: definition \:of\: logarithm}[/tex] :-
[tex]\ln|M-P| = -kt - C[/tex]
⇒ [tex] |M - P| = e^{-kt - C}[/tex]
⇒ [tex]M - P = \pm e^{-kt - C} [/tex]
[tex]\textbf{Further solving}[/tex] :-
⇒ [tex]M - P = \pm e^{-kt - C}[/tex]
⇒ [tex]M - P = \pm e^{- C - kt}[/tex]
⇒ [tex]M - P = \pm e^{- C + (- kt)}[/tex]
⇒[tex]M - P = \pm e^{- C} × e^{- kt}[/tex]
⇒ [tex]M - P = Ke^{- kt}[/tex]
⇒ [tex]M = Ke^{- kt} + P[/tex]
⇒ [tex]P = M - Ke^{- kt}[/tex]
[tex]\textbf{Let \:P(0) \:= \:0, \:then \:set\: t\: =\: 0\: and\: P \:= \:0}[/tex] :-
[tex]0 = M - Ke^{- k × 0}[/tex]
⇒ [tex] 0 = M - K × 1[/tex]
⇒ [tex]M = K[/tex]
Substitute th' following value in Original Eq :-
[tex]\boxed{P = M - Me^{-kt}}[/tex]
~ Hope it helps!
