# Complex differential equations to find functions [TRIED]

• NZBRU
In summary: A times [A/k / P + -A/k / (P + k/A)] dP= ∫-1/A times [A/k / P + -A/k / (P + k/A)] = ∫-1/A times [-1/A+1/Pk/A] = ∫-1/A times [0+(-1/A)] = ∫-1/A times [-1/A] = ∫-1/A times 0In summary, the homework statement is that two differential equations need
NZBRU

## Homework Statement

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

## Homework Equations

^As above, no other equations where provided, it is pretty much calculus and algebra

## The Attempt at a Solution

dP=kP-AP2 dt
dP=P(k-AP) dt
1/P dP = k-AP dt
∫1/P dP = ∫k-AP dt
Ln(P)=t(k-AP)
Ln(P)=tk-APt
etk-APT=P
loge(tk-APt)=P
eP=tk-APt

From there it just loops around. I can not think of anything else to do with it. As you can not (that I am aware of, unless you used unreals, factorise kP-AP2.

Thanks for reading this, I am not giving up and will continue to work on it now.

$$\frac{dP}{P(k-AP)} = \frac {-1} A \frac {dP}{P(P-\frac k A)}$$Use partial fractions.

Thanks a lot, I have now come down to: XP2- XPk/A - AY=1 (Instead of using A and B for partial fractions I used X and Y as those variables were not already used). The questions we have done in class involve something like: x-3=Ax-A+Bx+B and we just split them up into terms which contain x and terms not containing X to solve for A and B. Would you split it up into terms containing P and not P or terms not containing P or k or A?

We have not done this type of question containing two unknown variables (apart from those used for the partial fractions of course) yet. I apologize that I do now know how to use MathsJax, I am just using the quick symbols.

EDIT: Should I have used three fractions and used X,Y,Z as P(k-AP) could be factorized to:

-A times P times P - k/a. Then split the the final line into variables containing each one (somehow, some might contain two variables but I suppose they could be eliminated when finding the values for X, Y and Z). I will try it now...

Last edited:
Just call the P's x's and call -k/A "c" and you will feel right at home$$\frac 1 {x(x-c)}$$with a -1/A out in front.

That makes a lot more sense, I am doing it now. I assume the -1/A can be simply integrated to -Ln(A) as you just simplified -1/A to -1 times 1/A. I will post my results once I finish, thank you a lot by the way. I would be unable to do this without your help.

This is my working:

∫1/(Pk- AP2) dP
= ∫-1/A times p(p-k/a) dP
∫-1/A dP can be solved
= -1P/A (this could be incorrect)
now back to the other part (I know this setting out is terrible)
∫ p(p-k/a) dP

let P=x and -k/A=c

∫ x(x-c) dx
∫ A/x +B/(x-c) dx
∫ A(x-c)/x(x-c) +Bx/x(x-c) dx
∫ A(x-c)+Bx / x(x-c) dx
∫ Ax-Ac+Bx / x(x-c) dx

Therefore Ax-Ac+Bx / x(x-c) = 1
Split up into terms containing x and terms not containing x

X
A+B=0
B=-A

Other
-Ac=1
-A=1/c
A=-1/c

Therefore B= 1/c

-1/c / x + 1/c / (x-c) = 1 / Pk-AP2
Sub C and P back in...
A=-1/c
A=-1/(-k/A)
A =A/k

B=1/c
B=1/(-k/A)
B=-A/k

So you get:

∫-1/A times [A/k / P + -A/k / (P + k/A) ] dP

NZBRU said:
That makes a lot more sense, I am doing it now. I assume the -1/A can be simply integrated to -Ln(A)

Why would you assume that? -1/ A is just a constant out in front. The variable of integration is P (or x if you renamed it).

Oh, that makes sense. My bad, it was lake when I typed that. Is my working for the A and B values correct, I will try to differentiate the final line now.

## 1. What are complex differential equations and how are they used to find functions?

Complex differential equations are mathematical equations that involve both complex numbers and derivatives of functions. They are used to model many physical phenomena in fields such as physics, engineering, and economics. To solve these equations, one must find a function that satisfies the equation and its initial or boundary conditions.

## 2. What is the difference between a regular differential equation and a complex differential equation?

A regular differential equation involves only real numbers and derivatives of functions, while a complex differential equation involves complex numbers and derivatives of functions. Complex differential equations are more general and can model a wider range of phenomena.

## 3. Can complex differential equations be solved analytically?

Yes, some complex differential equations can be solved analytically using methods such as power series, Laplace transforms, and separation of variables. However, many complex differential equations require numerical methods to find solutions.

## 4. How do I know if a complex differential equation has a unique solution?

A complex differential equation has a unique solution if it satisfies the Cauchy-Riemann conditions, which ensure that the real and imaginary parts of the solution are continuous and have continuous derivatives. If these conditions are not satisfied, the equation may have multiple solutions or no solution at all.

## 5. Are there any real-world applications of complex differential equations?

Yes, complex differential equations have countless real-world applications, including modeling electrical circuits, fluid flow, heat transfer, and quantum mechanics. They are also used in economics, biology, and other fields to describe complex systems and make predictions.

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