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jonny [76]
3 years ago
9

Given: dp/dt = k(M- P)

Mathematics
1 answer:
arsen [322]3 years ago
5 0

Answer:

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

Step-by-step explanation:

The relation between the variables is given by

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

This is a separable differential equation. Rearranging terms:

\frac{dP}{(M- P)} = kdt

Multiplying by -1

\frac{dP}{(P- M)} = -kdt

Integrating

ln(P-M)=-kt+D

Where D is a constant. Applying expoentials

P-M=e^{-kt+D}=Ce^{-kt}

Where C=e^{D}, another constant

Solving for P

P=M+Ce^{-kt}

With the initial condition P=0 when t=0

0=M+Ce^{-k(0)}

We get C=-M. The final expression for P is

P=M-Me^{-kt}

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

Keywords: performance , learning , skill , training , differential equation

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