Basic Idea of Differential Equations

[NoahsArk]

TL;DR Summary

Why is this a subject of it's own.

Hello. After a lot of researching, I am still not clear how the subject of differential equations is really any different from derivatives and integrals which are learned in the main part of calculus. For example:

"Population growth of rabbits:

N = the population of rabbits at any time t
r= growth rate (.01 per week in the example)
$\frac {dN}{dt}$ = the population's rate of change

Think of $\frac {dN}{dt}$ as how much the population changes as time changes, for any moment in time.

Let us imagine the growth rate r is .01 new rabbits per week for every current rabbit. When the population is 1,000, the rate of change $\frac {dN}{dt}$ is then 1,000 x .01 = 10 new rabbits per week.

But that is only true at a specific time, and doesn't include that the population is constantly increasing. The bigger the population, the more rabbits we get!

When the population is 2,000, we get 20 new rabbits per week.

So it is better to say that the rate of change (at any instant) is the growth rate times the population at that instant:

$$ \frac {dN}{dt} = rN $$

And that is a Differential Equation, because it has a function, N(t) and its derivative." https://www.mathsisfun.com/calculus/differential-equations.html

What am I missing? Why does the equation $\frac {dN}{dt}$ have any strength to it, and how is this concept any different than something learned in the derivatives part of calculus? With derivatives, instantaneous rates of change, like the one in the rabbit example, are also discussed. The whole subject of differential equations seems to be reintroducing an idea taught at the beginning of calculus. I know I must be missing something fundamental given that all sources I looked at say how useful and important differential equations are.

btw- how can I preview a post before sending it? I noticed the "preview" tab now moved, but when I press it, nothing changes and I can't check if my latex is correct.

Thanks,

Joe

 

[jedishrfu]

With differential equations you can describe systems that have no clear analytic solution and that are far richer than a derivative of most functions. For some equations, only numerical solutions are possible.

In fact Differential Equation theory grew out of Calculus and so a decision was made by some teachers long ago to split Calculus into separate subjects.

 

[phyzguy]

When learning derivatives in calculus, you learned how to calculate dN/dt when you are given N(t). However, often you can write an equation like dN/dt = -rN, and you don't know the function N(t). Once you know the techniques of differential equations, you can solve for the function N(t) given the differential equation. Much of physics involves writing partial differential equations from your knowledge of the physics of the situation, then solving them to determine the functions you are interested in.

 

[fresh_42]

Summary:: Why is this a subject of it's own.

Hello. After a lot of researching, I am still not clear how the subject of differential equations is really any different from derivatives and integrals which are learned in the main part of calculus. For example:

"Population growth of rabbits:

N = the population of rabbits at any time t
r= growth rate (.01 per week in the example)
$\frac {dN}{dt}$ = the population's rate of change

Think of $\frac {dN}{dt}$ as how much the population changes as time changes, for any moment in time.

Let us imagine the growth rate r is .01 new rabbits per week for every current rabbit. When the population is 1,000, the rate of change $\frac {dN}{dt}$ is then 1,000 x .01 = 10 new rabbits per week.

But that is only true at a specific time, and doesn't include that the population is constantly increasing. The bigger the population, the more rabbits we get!

When the population is 2,000, we get 20 new rabbits per week.

So it is better to say that the rate of change (at any instant) is the growth rate times the population at that instant:

$$ \frac {dN}{dt} = rN $$

So far so good. And now integrate wolves, and buffaloes into your system! The more wolves, the fewer rabbits, fewer rabbits mean fewer wolves, but fewer wolves mean more rabbits, and so on.

Differential equations can describe such systems (Lotka-Volterra), or pandemics (SIR models). Existence, uniqueness, and dependency of initial values are questions in this field which put it closer to differential geometry than to analysis.

And that is a Differential Equation, because it has a function, N(t) and its derivative." https://www.mathsisfun.com/calculus/differential-equations.html

What am I missing?

The wolves.

Why does the equation $\frac {dN}{dt}$ have any strength to it, and how is this concept any different than something learned in the derivatives part of calculus? With derivatives, instantaneous rates of change, like the one in the rabbit example, are also discussed. The whole subject of differential equations seems to be reintroducing an idea taught at the beginning of calculus. I know I must be missing something fundamental given that all sources I looked at say how useful and important differential equations are.

btw- how can I preview a post before sending it?

Rightmost button in the editor box.

I noticed the "preview" tab now moved, but when I press it, nothing changes and I can't check if my latex is correct.

Thanks,

Joe

 

[NoahsArk]

Thank you for the responses:

When learning derivatives in calculus, you learned how to calculate dN/dt when you are given N(t). However, often you can write an equation like dN/dt = -rN, and you don't know the function N(t). Once you know the techniques of differential equations, you can solve for the function N(t) given the differential equation.

So, if I'm understanding you right, learning derivatives is about finding the derivative given a certain function, and learning differential equations is about finding the function when you are given the derivative? How, though, is that different from integration. In integration, we start with a derivative and find the function that the derivative came from.

In the example of $y = x^2$ where y is position and x is time, we could write a differential equation saying $\frac {dy}{dx} = 2x$ From there we could use integration to get the function $y = x^2$ without knowledge of differential equations. Please let me know what I'm missing.

 

[phyzguy]

So, if I'm understanding you right, learning derivatives is about finding the derivative given a certain function, and learning differential equations is about finding the function when you are given the derivative? How, though, is that different from integration. In integration, we start with a derivative and find the function that the derivative came from.

In integration, you are still given the function that you are integrating. In solving a differential equation, you are not given a function. All you are given is a relation between the derivative and the function, like dN/dx = -r N. You have no knowledge of the function N. If you think about it, you will see that it is very different.

 

[NoahsArk]

So far so good. And now integrate wolves, and buffaloes into your system! The more wolves, the fewer rabbits, fewer rabbits mean fewer wolves, but fewer wolves mean more rabbits, and so on.

Differential equations can describe such systems (Lotka-Volterra), or pandemics (SIR models). Existence, uniqueness, and dependency of initial values are questions in this field which put it closer to differential geometry than to analysis. The wolves.

Rightmost button in the editor box.

How could something like wolves be incorporated into the differential equation? Thanks

 

[fresh_42]

You have had rabbits. Wolves eat rabbits, so both populations are dependend on each other. And the more complex a system is, the more equations are necessary to describe it. Those equations are usually not independent.

Here are two examples:
https://www.physicsforums.com/threads/math-challenge-june-2019.972824/ - challenge no. 6
https://www.physicsforums.com/threads/math-challenge-november-2020.995557/ - challenge no. 8

The solutions - if not in the thread - are here:
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/

 

[Office_Shredder]

Differential equations are the same thing as integrating, only more complicated. Every integration is just solving $y'(x)=f(x)$. But what about $y'(x)=y(x)$? What about $y'(x)^2+2y(x)+\sin(y''(x))=0$?

 

[NoahsArk]

Differential equations are the same thing as integrating, only more complicated. Every integration is just solving y′(x)=f(x). But what about y′(x)=y(x)?

That's a helpful way to think about it. I'm confused on the notation though: $y\prime(x) = f(x)$
This doesn't have an integration symbol so I'm trying to see how integration is involved. How would this be written if the equation were $y = x^2$ ? Here, would $y\prime(x) = 2x$ ? Would f(x) be $x^2$ ? If so, it looks like a derivative problem and not an integration problem. Also, $y\prime(x)$ would not = f(x) since $y\prime(x) = 2x$ and f(x) = $x^2$ Please clarify.

Also, I'm trying to thinking of what something like y′(x)=y(x) means? Can you give an example?

Thanks

 

[fresh_42]

That's a helpful way to think about it. I'm confused on the notation though: $y\prime(x) = f(x)$
This doesn't have an integration symbol so I'm trying to see how integration is involved.

Integration is involved as the way to solve the equation. We have $y'=f$ and we need to integrate both sides to get $y=F$.

How would this be written if the equation were $y = x^2$ ? Here, would $y\prime(x) = 2x$ ? Would f(x) be $x^2$ ?

No. $f(x)$ would be $2x$. We would have $y'(x)=2x$ and looking for $y(x)=x^2+C.$

If so, it looks like a derivative problem and not an integration problem. Also, $y\prime(x)$ would not = f(x) since $y\prime(x) = 2x$ and f(x) = $x^2$ Please clarify.

Also, I'm trying to thinking of what something like y′(x)=y(x) means? Can you give an example?

Thanks

Let $y(x)=e^x$. Then $y'(x)=e^x=y(x)$. So $y(x)=e^x$ is a solution for $y'=y$. The theory of differential equation asks e.g. whether this is the only solution. It is not, because $y(x)=e^{x+C}$ is also a solution for $y'=y$. If we require an initial value, e.g. $y(0)=e^2$ then we get a unique solution because it fixes all possible values of $C$ to $C=2.$

This is the most elementary example of a differential equation. Now imagine if $y$ and/or $x$ were vectors, or we had many such equations simultaneously to solve. Things quickly get complicated. And that's why it is an extra course next to calculus.

Differential equations are actually the tools with which we describe nature.

 

[Delta2]

Basically differential equations are part of calculus and use techniques and theorems from both integral and differential calculus as well as other branches of math (e.g. algebra of vector spaces). But this part of calculus is so big on its own that it is studied as a separate branch of math.

The solution to any differential equation is expected to be some sort of integral, or to be found via integrations, however sometimes the differential equation is so complex that we don't know what exactly we should perform integration upon, or even if we setup the integral we can't perform the integration because there is no closed form for this integral (for example check elliptic integrals).
In both of the above cases we can try to find an approximation to the solution of the differential equation by using algorithms and techniques from numerical analysis.

 

[NoahsArk]

But what about y′(x)=y(x)? What about y′(x)2+2y(x)+sin⁡(y″(x))=0?

So, are all D.E.s equations where the variable we are trying to solve for is in the derivative form? E.g. $y \prime =$ something?

With integration we're trying to find the area underneith a curve. What are we trying to do with D.E.s?

 

[fresh_42]

So, are all D.E.s equations where the variable we are trying to solve for is in the derivative form? E.g. $y \prime =$ something?

With integration we're trying to find the area underneith a curve. What are we trying to do with D.E.s?

E.g. predict the size of populations, or estimate the course of a pandemic. See posts #4,#8.

 

[NoahsArk]

I meant more what are we trying to as can be illustrated on a graph? I'm trying to compare with regular integration problems to get the idea. If, for example, with integration we are trying to find the area under the curve of some function, say y = 2x, at a certain x value, then what are we trying to do by comparison with a D.E.?

 

[Delta2]

So, are all D.E.s equations where the variable we are trying to solve for is in the derivative form? E.g. $y \prime =$ something?

With integration we're trying to find the area underneith a curve. What are we trying to do with D.E.s?

If the D.E equation contains $y'$ then I suppose we can bring it to the form $y'=something$ but that "something " might or not contain y and some of its higher derivatives, e.g. the ODE $$y''+\sin x\cos y+y'=\tan x$$ can be brought to the form $$y'=\tan x-\sin x\cos y-y''$$ but as you can the "something" contains the unknown function y, and also its second derivative y'' so you can't integrate it directly to recover y as some function of x.

Trying to find the solution to an ODE is trying to find an integral so that the unknown function is equal to that integral. That integral will contain somehow the known functions (in the above example the known functions are $\tan x$ and $\sin x$ as well as some other functions), so it will be $$y=\int R(\tan x,\sin x,...)dx$$ where R is a proper expression depending on the form of the ODE, for example it might be $$R(\tan x,\sin x)=\sin^2x\frac{\tan x-1}{\tan x+\sin x}$$
P.S (I think the ODE given in the above example doesn't have a closed form solution).

 

[NoahsArk]

To try and use a basic example, take $\int 5x$. This is the same as writing $\frac {dy}{dx} = 5x$ right? In both cases the answer is $\frac {5x^2}{2}$? I am assuming this is a basic integration problem and not a D.E.. When we change the problem a little bit to $\frac {dy}{dx} = 5x-y$, that addition of a -y turns it into a differential equation?

It's hard for me to explain what I'm most stuck on, but the best way I can describe it is that I don't understand what the significance could be of adding a -y (or a 5y or any other term with a y in it? IOW what situation could we be trying to model by doing so and what does it even mean to be adding a -y? If we, for example, added an $x^2$ instead of a -y, it would still be a normal integration problem and not a D.E. By adding something like an $x^2$ we'd get $\frac {dy}{dx} = 5x + x^2$, and all the addition of the $x^2$ does is to add more area under the curve by increasing the value of the function. Adding a y term doesn't make sense to me, and it gets even more confusing to me if we start taking about adding something like a second derivative of y into the picture.

 

[fresh_42]

Both are differential equations, $y'=5x$ and $y'=5x-y$. They are only of different degrees of difficulty.

The major point is that a course in differential equations deals with far more complicated equations, simultaneous equations, and dependencies all over the place; neither of which is covered by simple differentiation.

You want to see an area? What if a tank of a certain volume is filled with water, that does not uniformly spill into the tank, but has varying pressure and flow over time. Think further that you may change the flow twice. How long does it take to fill the tank?

This is still an integral: water in a tank. But how do you model pressure, flow, time, and manual changes?

 

[NoahsArk]

Both are differential equations, y′=5x and y′=5x−y. They are only of different degrees of difficulty.

So I am assuming the answer to the second problem is a function whose derivative is 5x-y? I remember learning about these in the "implicit differentiation" section on derivatives.

 

[Mark44]

So, are all D.E.s equations where the variable we are trying to solve for is in the derivative form? E.g. y′= something?

As has already been said in this thread, a differential equation is an equation involving an unknown function and one or more of its derivatives. There are two main subdisciplines: ordinary differential equations (ODE), in which the functions are of a single variable; and partial differential equations (PDE), in which the functions are of two or more variables.

With integration we're trying to find the area underneith a curve.

That is only one application of the definite integral. There are many more.

 

[bob012345]

So I am assuming the answer to the second problem is a function whose derivative is 5x-y? I remember learning about these in the "implicit differentiation" section on derivatives.

What we are trying to find is simply the relationship between $y$ and $x$ or $y = f(x)$.

In the first case $y=\frac{5x^2}{2}+c_1$ and in the second case $y=5\left(x-1\right)+\frac{c_1}{e^x}$ where $c_1$ is an arbitrary constant of integration.

Doing these by direct integration, the first is easy $\frac{dy}{dx}=5x$ gives $∫dy = ∫5xdx$ or $y=\frac{5x^2}{2}+c_1$.

Integrating the second case $\frac{dy}{dx}=5x - y$ we get $∫dy = ∫5xdx - ∫ydx$

which gives us $y=\frac{5x^2}{2}+c_1 - ∫ydx$ since we do not know how $y$ depends on $x$ we are stuck. We could guess what $y = f(x)$ is and solve it by trial and error but the field of Differential Equations simply is the compilation of all the tricks and tools to figure that relationship out and get us unstuck.

 

[PeroK]

I meant more what are we trying to as can be illustrated on a graph? I'm trying to compare with regular integration problems to get the idea. If, for example, with integration we are trying to find the area under the curve of some function, say y = 2x, at a certain x value, then what are we trying to do by comparison with a D.E.?

Differential equations are fundamental to physics. We can start with Newton's second law, where (in general) the force on a particle is a function of time: $$F(t) = ma(t)$$ If we note that acceleration is the second derivative of position, then we have: $$m\frac{d^2x}{dt^2} = F(t)$$and we see that Newton's second law is a second-order differential equation. Solving this differential equation gives us the position of a particle for a given force.

Classical EM (Electromagnetism) is governed by Maxwell's equations, which are a set of four partial differential equations relating the electromagnetic field to electric charges and currents. If we know the charges and currents, then solving Maxwell's equations gives us the Electromagnetic field.

If we move on the QM (Quantum Mechanics) we find Newton's laws are replaced with the Schroedinger equation (SDE): $$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V\Psi$$where $\Psi(x, t)$ is the wave-function and $V(x, t)$ is the potential. The SDE is a second-order partial differential equation, as it has both a time derivative and a second-order spatial derivative. Solving the SDE is how we determine the behaviour of a QM system governed by a given potential.

In general, therefore, the laws governing physical systems are often encapsulated in one or more differential equations. And, solving those equations tells us about the behaviour of the physical system in question.

 

[mathwonk]

geometers think of diff eqs in terms of vector fields. functions of various sorts have derivatives which can be viewed as vector fields, and conversely, given a vector field, one can ask whether they arise from differentiating certain special types of functions.

for instance a smooth function defined on the plane has a gradient field, assigning to each point of the plane the gradient of the function at that point. The inverse problem is to look at a vector field in the plane and ask whether it arises as a gradient field for some smooth function, and if so, to find that function.

The solution is found by considering the path integrals of the vector field taken along curves in the plane. By the Fundamental theorem of calculus, the integral of a gradient field along some curve equals the difference in the values of that function at the ends of the path. In particular the integral in this case depends only on the end points of the path and not on the particular trajectory taken by the path in going from the initial point to the terminal point. Indeed, any vector field which has this property, that path integrals over it depend only on endpoints of the paths, is a gradient field, and the function whose gradient the field is, can be computed, up to a constant, by path integration. I.e. since the integral of df along a path from p to q equals f(q)-f(p), given the integral and fixing an arbitrary choice of the value f(p), we can use this formula to define f(q) for every q. The condition of being a gradient is detected locally by the vanishing of the “curl” of the vector field. I.e. a vector field whose curl is zero, is a gradient at least within any disc contained in its domain, but if we range over the entire domain, the function of which it is a gradient may not be well defined on a loop that encloses a point outside the domain. e.g. the vector field dtheta, is defined and has curl zero everywhere away from the origin, but the angle function theta cannot be continuously defined on any loop containing the origin.

This technique gives a fascinating way to study the topology of smooth surfaces. Visualizing a g-holed doughnut say, standing on end, the height function has one max, one min, and 2g saddle points on the doughnut. Thus the gradient field will have 2g + 2 zeroes, since the gradient at any critical point is zero. But using the matrix of second derivatives we can distinguish extrema from saddle points, and assign a sign to these critical points, +1 to a max or min, and -1 to a saddle point. Then the sum of the signed values assigned to these critical points is 2-2g, exactly the euler characteristic of the surface! In fact any general vector field can be used to calculate the euler characteristic, gradient or not. A converse is also true and enables one to classify compact 2 manifolds. Namely on such a manifold (say oriented) there always exists a smooth function with one max one min and a finite number 2g of saddle points. Using flows arising from solving (as in the next paragraph) the differential equation associated to the gradient field, it follows the manifold is homeomorphic to a doughnut with g handles. Even in higher dimensions, a compact oriented manifold with a smooth function having exactly two critical points, is homeomorphic to a sphere, proved by the same method.

Since a more general vector field in the plane usually will not satisfy the criterion of path independence for integrals, we seek a different way to realize them geometrically. Recall that a smooth path in the plane will have a velocity vector at each point. Given a smooth vector field in the plane and a point, we can ask whether there is a path through that point whose velocity vectors agree with the specified vectors at least at points of that path. Not only is this usually true, but it can be done with a whole family of paths, so that the velocity vectors of all the paths in fact fill up a region in the plane and exhaust all the given vectors there.

Thus given a vector field in a region in R^n, we have ways to find either a family of maps R—>R^n, i.e. paths, or in some cases a single map R^n—>R, i.e. a function, whose derivatives give back the vector field. And as discussed above, looking at the "family of paths" solution to a field originally defined as a gradient, illuminates the topology of the manifold.

i have not studied this but just from trying to explain it here the idea seems to have become clear. Imagine a compact 2 manifold with a smooth function having exactly two critical points, both “non degenerate”. Then look at the gradient field of this function. The critical points must be a max and a min, and since they are non degenerate, the gradient vectors all point in radially at the max and all point out radially at the min. Now draw a little disc around each critical point and remove these 2 discs from the manifold. We are left with a manifold with boundary, whose boundary consists of two circles, and a gradient vector field on it which is everywhere non zero, with vectors all pointing away from the min circle and all pointing in toward the max circle. Now fill in the “flow” or path solutions to this differential equation, hence filling the manifold by paths that extend from the lower circle to the upper circle, all disjoint from each other, i.e. parallel, and filling up the surface. What is this surface? It seems persuasive that it is a cylinder. Then what was the original manifold, obtained by replacing the two discs, one at the top and one at the bottom. Surely this is a sphere.

The same argument applies to classify any oriented surface, but we must remove in addition to the two discs at the max and min, also one saddles at each of the other critical points. It is not as easy to picture, but we again obtain for the complement a disjoint union of cylinders, and when we replace not only the two discs, but also the saddles, one can show we get a doughnut with half as many holes as saddles. E.g. a single holed torus is the union of two discs and two saddles and two cylinders. This called “Morse theory”.

Notice how clever is the method: we start from a differential equation whose solution we know as a gradient. I.e. we start from a function, and look at its gradient field. So we know the solution to the equation asking us to find a function whose gradient gives this field. But then we flip the question around and ask instead for a family of paths solving this same differential equation. That solution also exists and gives us a description of the manifold in terms of paths, which helps us better understand the manifold. I.e. giving a function f:M-->R on our surface M, decomposes M as a union of level curves: for each constant a, we have the curve of {all x in M : f(x) = a}. But then if we look at the gradient field of this f, and fill up this gradient field by paths, by get a description of M in terms of maps R^2-->M, which gives a better description of the manifold by a union of "charts".

There are also criteria for determining whether the family of paths satisfying a vector field defines a coordinate system in the plane, in terms of the vanishing of the “lie bracket” of the vector fields, which basically reflects the fact that mixed partials are equal for a function of several variables. I.e. a solution to the problem of realizing a vector field as the velocity vectors of a family of paths, is essentially a map F:R^2—>R^2, such that through each point p = (F(x0),F(y0)), the solution path is the restriction of F to the line (x,y0) obtained by fixing y = y0, and letting x vary. This means the vector field is given by the family of velocity vectors ∂F/∂x. But this function F also defines another velocity vector field, namely ∂F/∂y. So a more refined problem is to give ourselves two vector fields in the plane and ask when there exists a "change of coordinates" map F such that the first vector field is given by ∂F/∂x and the second is given by ∂F/∂y. The equality of mixed partials of F forces in this case the vanishing of the so called :”bracket product” of the two given vector fields, and conversely that is a sufficient condition for the existence if such a solution map F. i hope i got this more or less right, but i am a novice in diff geom.

People are also interested in other partial diff eq’s, and these too can have a geometric interpretation. Laplace’s equation, for instance comes up famously in complex analysis, as satisfied by the real and imaginary parts of holomorphic functions. It also yields, in complex geometry, a refinement of the famous de Rham theory from differential topology, expressing the cohomology of a smooth manifold by means of smooth differential forms. In this theory, which is analogous to that detecting when a vector field is a gradient, each cohomology class is represented by a whole coset of “closed” differential forms, analogous to those with curl zero, modulo the subspace of “exact” forms, those analogous to gradients. But in the presence of a complex structure on our manifold, we can also define “harmonic” forms, those satisfying the Laplace equation. It turns that then in each cohomology class there is a unique closed harmonic form representing that class, modulo exact forms. It is obviously more convenient to be able to work with a specific form rather than a whole coset of them, but I am not an expert.

My own special favorite is the heat equation, for a function F(t,z) of several complex variables, (tjk, zi), 1≤ i,j,k ≤ n, which equates the first derivative of F wrt tjk, to a constant multiple of the second derivative wrt zj,zk. The fundamental solution function F, the famous “theta function” of Riemann, defines a family of hypersurfaces, defined by fixing t and setting F(z,t) = 0 as a function of z, where these hypersurfaces live in a family of n dimensional complex analytic tori, or abelian varieties. The geometry comes from the heat equation, which links the quadratic tangent cones at double points of these hypersurfaces, to tangent hyperplanes in the family parametrizing these complex tori. The most famous such tori are those defined as jacobian varieties of complex curves, or Riemann surfaces, and using the heat equation, one can show that the locus of such jacobians in the moduli space of tori, agrees locally with the locus where the given hypersurface has a fixed dimensional locus of double points. One can also show that the given Riemann surface can in general be cut out in the projective space defined by lines in the tangent space at the origin of the torus, by those quadric hypersurfaces that occur as tangent cones to double points of the hypersurface defined by the theta function, i.e. the curve is determined by its jacobian variety, Torelli’s problem.
any interested party may consult this link:
http://www.numdam.org/item/CM_1990__76_3_367_0.pdf

for more on the geometry of differential equations i strongly recommend any book by v. arnol'd, such as his intro to ordinary diff eq.

https://www.amazon.com/dp/0262510189/?tag=pfamazon01-20

another undergraduate source for using differential equations to do topology is Wallace:

https://www.amazon.com/dp/0486453170/?tag=pfamazon01-20

 

[mathwonk]

to the original poster. just a throwaway, but your question about why a general differential equation like f'''' + af'''+bf'' + cf' + df = 0, is any different from integrating to find f when f' = g, is a little like asking why an algebraic equation like x^5 + ax^4 + bx^3 + cx^2 + dX + e = 0, is any different from a linear equation ax = b. Namely it is just not as obvious how to solve for f, or x.

if you want to take a leap to the higher realms of math, you might also take a peek at my previous post, but I apologize if that is a kick in the head. But in any case pay no attention to the numdam link, which is for grad students and higher.

 

[NoahsArk]

I appreciate the responses.

 

[Astronuc]

A simple example can be found with radioactive decay.
http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli.html
http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/halfli2.html#c1

It's simple when one is concerned with a radionuclide that decays into a stable product. It gets more complicate when one radionuclide decays into another radionuclide, or there is a nuclear reaction that is creating a radionuclide, which is decaying.

Then think about a nuclear reactor in which fission creates dozens of radionuclides, which not only decay, but some are transformed due to neutron absorption.

Differential equations, and systems of differential equations, are found in control theory and many other practical applications having to do with fluid flow, heat transport, chemical reactions, rocket propulsion, spacecraft trajectories, . . . .

Summary:: Why is this a subject of it's own.

When the population is 2,000, we get 20 new rabbits per week.

With respect to the rabbit problem, others have mentioned the effect of predators (not only wolves, but bobcats, eagles, hawks, . . . ). There is also the effect of food. Rabbits will reproduce as long as there is food. If the food supply decreases, the rabbit population will decrease. And then there is disease, and closer to the polar region, harsh winter weather.