Very complex differential equations to find functions [TRIED]

[nzbru]

I need to solve two differential equations to find a population function P(t). I am able to do this with problems like newtons law of cooling:

dT/dt=-k(T-Ta) solves to:
dT=-k(T-Ta) dt
∫1/(T-Ta)dT=∫-k dt
Ln(T-Ta)=-kt
e^-kt=T-Ta
T=Ta+E^-kt

However I have been presented with two hard problems which I can not do:

1. dP/dt=kP-AP²
2. dP/dt=kP(M-P)

using P=P_o at t=0 find P(t) for both.

I have mostly completed the first one and have now got it simplified to:

e^{t+c =} KP(K²-APK)^A where C = In(KPo(K²-APoK)^A) [NOTE: Po is another variable which was introduced by using the logic when t=0 P must equal its initial value].
Now I am unable to find an expression for P. I have tried multiple times and I cannot find a way to work around the brackets being to the power of A.

May someone help me or guide me in the right path of how to find an expression for P as I am stuck. Cheers, NZBRU.

 

[HallsofIvy]

I need to solve two differential equations to find a population function P(t). I am able to do this with problems like newtons law of cooling:

dT/dt=-k(T-Ta) solves to:
dT=-k(T-Ta) dt
∫1/(T-Ta)dT=∫-k dt
Ln(T-Ta)=-kt
e^-kt=T-Ta
T=Ta+E^-kt

However I have been presented with two hard problems which I can not do:

1. dP/dt=kP-AP²

So \(\displaystyle \frac{dP}{kP- AP^2}= \frac{dP}{P(k- AP)}= dt\)
Use partial fractions to write this as
\(\displaystyle \frac{dP}{P}+ A\frac{dP}{k- AP}= kdt\) and, integrating,
\(\displaystyle ln(P)- ln(k- AP)= kt+ C\)
Taking the exponential of both sides \(\displaystyle \frac{P}{k- AP}= e^{kt+ C}= c e^{kt}\(\displaystyle
where \(\displaystyle c= e^C\). With \(\displaystyle P(0)= P_0\), that gives \(\displaystyle \frac{P_0}{k- AP_0}= c\)
But I don't get anything like what you have below. In particular, I can't see where you got the "A" power.

2. dP/dt=kP(M-P)

Similarly this can be written
\(\displaystyle \frac{dP}{P(M- P)}= kdt\) and then
\(\displaystyle \frac{1}{M}\frac{dP}{P}+ \frac{1}{M}\frac{dP}{M- P}= kdt\)

using P=P

_o at t=0 find P(t) for both.

I have mostly completed the first one and have now got it simplified to:

e^{t+c =} KP(K²-APK)^A where C = In(KPo(K²-APoK)^A) [NOTE: Po is another variable which was introduced by using the logic when t=0 P must equal its initial value].
Now I am unable to find an expression for P. I have tried multiple times and I cannot find a way to work around the brackets being to the power of A.

May someone help me or guide me in the right path of how to find an expression for P as I am stuck. Cheers, NZBRU.

\)\)

 

[nzbru]

From dP/dt=kP-AP²
I too s=used similar fractions but used this working:

1/(P(k-AP)) dP= 1 dt
t= X/P + Y/(k-AP)

So [FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]t[/FONT]
This step I understand but the next one I do not get
I will try re-trying the partial fractions section now.
[FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]t[/FONT]


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[nzbru]

I have done it by by leaving the dP on the top of the fraction and have gotten to:

dP=xk-APx+YP

Can you assume because of the dP is multiplied by one that dP=P which would make the next line:

P=xk-APx+YP and then could split the components containing X and not containing X apart.