# What is a differential?

1. Jan 21, 2005

### LENIN

What is a differential???

I have a strange problem. I am supose to enter a physics cometition in about 2 months, where the knowledg of differential equations is almost necasery, but we haven't done yet any at our mathemathics or physics class. Our mathemathics teacher doesn't wan't to explein them to me and the books we are useing also have almost no explination. I would be very happy if someone could plees explain the basics. Thanks.

2. Jan 23, 2005

### drgnclwk

this from my text book:
1. differentiate both sides of the equations with respect to x
2. collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation.
3. factor dy/dx out of the left side of the equation
4. solve for dy/dx by dividing both sides of the equation by the left-hand factor that does not contain dy/dx

so example:
find dy/dx given that y^3+y^2-5y-x^2=-4
step 1
d/dx[y^3+y^2-5y-x^2]=d/dx[-4]
d/dx[y^3]+d/dx[y^2]-d/dx[5y]-d/dx[x^2]=d/dx[-4]
3y^2dy/dx+2ydy/dx-5dy/dx-2x=0
step 2
3y^2dy/dx+2ydy/dx-5dy/dx=2x
step 3
dy/dx(3y^2+2y-5)=2x
step 4
dy/dx=2x/(3y^2+2y-5)

p.s. how mean, ur math teacher doesn't want to show you how to do math???

Last edited: Jan 23, 2005
3. Jan 23, 2005

### dextercioby

Out there,there are many types of Diff.Eq.We cannot explain to you the basics of their theory because it is too long...In the prior post,the "author" gave a rather lucrative,yet unrigurous treatment of the theorem of implicit functions.That has nothing to do with differential equations.
I don't know what to say.I cannot foretell what sort of equations u'll be encountering at the competition.It may be simple ones (i-st order linear) or very complicated (higher order unlinear).

Give us a specific example and we'll show u how to do it.

Daniel.

4. Jan 23, 2005

### arildno

Simplest put:
Differential equations are a type of functional equations (i.e where the unknown is some function) which explicitly contain derivatives of the sought function.

5. Jan 23, 2005

### drgnclwk

sorry... i was thinking he needed to differentiate equations...
if you're solving for a differential equation...
dy/dx=f(x)
dy=f(x)dx
then integrate

y=∫f(x)dx=F(x)+C
(F(x)=antiderivative, C=constant)

so for simple example

dy/dx=√(x)+x^2-4x^3+2
dy=(x^(1/2)+x^2-4x^3+2)dx

to get the antiderivative add 1 to your exponent then divide the variable by your new exponent

y=∫(x^(1/2)+x^2-4x^3+2)dx
y=(2/3)x^(3/2)+(1/3)x^3-x^4+2x+C

dunno, maybe this is something you can use..

6. Jan 24, 2005

### Galileo

7. Jan 30, 2005

### LENIN

Thanks for all yor help!!!

8. Jan 30, 2005

### mathwonk

In a differential equation, you are given some equation satisfied by some derivatives of the function, and asked to find the function. Since there are usually many solutions, for example if f'(x) = 0, any constant will do, you are also usually given some particular values of the functions and some of its derivatives.

E.g. here is anice d.e. f' = f, and f(0) = 1. This is only satisfied by f(x) = e^x. All you need to prove this is to know that the only solution of f' = 0 is a constant.

I.e. if f' = f, then the derivative of f/e^x is zero by the quotient rule. so f/e^x is a constant c, so f(x) = ce^x, and since f(0) = 1, we get c = 1.

similarly, if f' = af, then f =c e^(ax), by the same argument.

This simple result leads to the theory of linear constant coefficient differential equations as follows:

suppose we write the equation f' = af, as f'-af = 0, or (D-a)f = 0, where D is differentiation and a is multiplicatioin by a. then we know all solutions of (D-a)f = 0 from the previous argument.

Now suppose we consider a second order de of this type: say f'' - f = 0, written as (D^2 - 1)f = 0. Then we factor the operator as D^2 -1 = (D-1)(D+1), and we see that, since these operators commute, any solution of either factor is a solution of their "product", just as with algebraic equations.

I.e. both e^x and e^(-x) are solutions, as is then any linear combination of form ae^x + be^(-x).

We claim these are the only solutions. this is just, linear algebra.

i.e. if the product (D-1)(D+1)f = 0, then g = (D+1)f must solve (D-1)g = 0, hence g must equal ae^x. So the only solutions f of (D-1)(D+1)f = 0, are solutions f of (D+1)f = ae^x. Well, the only solutions of this are of form: particular solution plus general solution of (D+1)f = 0. Since it is easy to see that f = (a/2)e^x solves
(D+1)f = ae^x, we get that the general solution of (D+1)f = ae^x, has form

f = be^(-x) + (a/2)e^x. Thus the general solution of (D-1)(D+1)f = 0 is ae^x + be^(-x).

i.e. the solutions are just linear combinations of the solutions of the factors taken separately: (D+1)f = 0 and (D-1)f = 0.

Similarly, any d.e. of form (D-a)(D-b)....(D-c)f = 0, where all the roots a,b,...,c are distinct, has as solutions only linear combinations of the functions e^(ax), e^(bx),....,e^(cx).

So we look at the equation (aD^n + bD^(n-1) +.....+cD + d)f = 0, and form the associated algebraic equation aX^n + bX^(n-1) +....+cX + d = 0 and solve it.

If it factors into distinct linear factors, then this process gives all the solutions. If not there is a little twist that does give them.

More general differential equations are very hard, but everyone should know this much, i.e. how to solve the easiet ones.

Then a physicist should also learn the three most interesting classical differential equations of more variables ("Partial d.e.'s"), the laplace equation, the heat equation, and the wave equation.

the laplace equation is essentially ∂^2f/∂x^2 + ∂^2f/∂y^2 = 0, and its solutions are called harmonic functions. they form the real parts of complex differentiable functions. there is also a version in more variables.

the heat equation is essentially ∂f/∂tab = c ∂^2f/∂xa∂xb, where f(x,t) is a function of x and t, where x has n coordinates and t has "n+1 choose 2" coordinates.

the fundamental solution is called a "theta function". these can be written out in terms of fourier series. they play a big role in Riemanns theory of complex curves.

I am not familiar with the wave equation.

Last edited: Jan 30, 2005
9. Jan 30, 2005

### Chronos

Your teacher probably resists explaining DEQ [differential equations] because he/she does not understand them - which is not uncommon for a secondary school teacher [you don't have to understand DEQ to teach at that level].

10. Jan 31, 2005