Solving the Diff. Equation: P(t) = (MP_{0}) / (P_{0} + (M - P_{0})e^{-kMt})

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The discussion revolves around solving the differential equation dP/dt = k(M - P)P and verifying that the solution matches the form P(t) = (MP_{0}) / (P_{0} + (M - P_{0})e^{-kMt}). Users share their attempts at integrating and manipulating the equation, expressing confusion about the presence of three terms in the denominator of the expected solution. There is a focus on the integration process and the correct application of initial conditions, particularly regarding the constant of integration. Participants also discuss the challenges faced in reaching the correct form and the potential errors in their methods. The conversation highlights the complexities of solving differential equations and the importance of careful algebraic manipulation.
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Homework Statement



Solve the euqation dP/dt = k(M - P)P to show that it equals: P(t) = (MP_{0}) /( P_{0} + (M - P_{0})e^{-kMt})

Homework Equations



The Attempt at a Solution


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

\frac {1} {M}\ \ln( \frac {P} {M - P}) + C = Kt + C

I combine the constants of integration... i can do this right?
Then i get rid of the log by taking the exponent of both sides.

P/(M - P) = e^{mkt + c}

I then turn e^{c} into P_{0}

Next i divide through by (M - P)

P = (M - P)P_{0}e^{mkt}

next i combine the P then divide through by the remainder.

P = (mP_{0}e^{mkt})/(1 + P_{0}e^{mkt})

and i end up with:
P = (mP_{0})/(e^{-mkt} + P_{0})

In the equation i am suppose to get i don't see how there could possibly be three terms in the denominator.
 
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You could just differentiate P and substitute it into the differential equation.
 
thanks for the reply cristo.

I was in the process of editing my post (i am still getting use to LaTex) when you replied. i could do that but my teacher want us to just solve it. Also this has been bugging me for a while, i have been working on it for the past few hours and cannot get it. My teacher... well a grad student TA for my DE lab couldn't solve it either.

I also would like to know what is wrong with my method... it seems like it should work but... its not.

one thing think might be wrong with my method is the P_{0} initial condition.
 
dionysian said:

Homework Statement



Solve the euqation dP/dt = k(M - P)P to show that it equals: P(t) = (MP_{0}) /( P_{0} + (M - P_{0})e^{-kMt})

Homework Equations



The Attempt at a Solution


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

\frac {1} {M}\ \ln( \frac {P} {M - P}) + C = Kt + C

I combine the constants of integration... i can do this right?
Then i get rid of the log by taking the exponent of both sides.

P/(M - P) = e^{mkt + c}

I then turn e^{c} into P_{0}
HOW did you "turn e^{c}[/tex] into P_0&quot;? <br /> <br /> When t= 0, You have P_0/(M-P_0)= e^c<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Next i divide through by (M - P)<br /> <br /> P = (M - P)P_{0}e^{mkt} <br /> <br /> next i combine the P then divide through by the remainder. <br /> <br /> P = (mP_{0}e^{mkt})/(1 + P_{0}e^{mkt})<br /> <br /> and i end up with:<br /> P = (mP_{0})/(e^{-mkt} + P_{0})<br /> <br /> In the equation i am suppose to get i don&#039;t see how there could possibly be three terms in the denominator. </div> </div> </blockquote>
 
Thanks a million. works out fine after that.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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