Looking at Basic Differential Equations (1)

As I pointed out in my blogs on quantum mechanics(QM), differential equations are critical to understanding and using QM. Indeed, the basic foundational QM equation is the Schrodinger equation, a partial differential equation of order two. But, before one can "run" with the higher order partials, one must be able to recognize and solve the humdrum, ordinary linear differential equations. So in the next few instalments (spaced apart!) we will look at different aspects.

As readers may be aware, math is definitely a part of this blog and from time to time I will veer into one or other aspect of mathematics. For those who'd like a compilation of all the math blogs thus far, you can obtain my book: Mathematical Excursions in Brane Space, which is now available at lulu.com. The book is available both in e-format (file download) for $5.00 or as a paperback, 156 page book, for $9.50.

For the sake of reducing the background info, I'm assuming all readers who venture into this area have had at least some exposure to differential calculus. If not, or if you're rusty, you can brush up here:

http://en.wikipedia.org/wiki/Differential_calculus

Let's start our venture into DEs with just about the simplest one imaginable:

dy = x dx

As with all DEs, the solution is accomplished via the process called integration.

(see, e.g. http://en.wikipedia.org/wiki/Integral)

If we integrate both sides, we obtain:

y = x^2/ 2 + c

where c is some undefined (as yet) constant of integration. We call the above the "general solution" to the differential equation. This general solution is, in fact, a family of parabolas. A table of values can also be calculated here to expose the limits for x, y entries. (Readers can do this starting with x = 0 and for corresponding dy/dx (or y'). Thus, for x = 0, y' = 0, and for x = +1/2 (or -1/2) then y' = +1/2 (or -1/2) and for x = +1 (or -1) then y' = +1 (or -1) and so forth. )

What the reader will find when all is over and done with is a graph family such as shown. If we wanted to obtain the particular solution, we'd have to have boundary conditions available. Usually these designate what values x, y are to have at a particular point, and also often the first derivative (y' or dy/dx) at the same point.

Thus, we enter the world of first order differential equations of the first degree.

Let's consider the equation:

xdx + ydy = 0

One could be understandably tempted to write this in terms of the 1st derivative to obtain:

dy/dx = -x/y

but this serves no useful purpose. It's more productive to simply integrate the equation:

xdx = -ydy

directly (since variables are already separated) to obtain:

x^/2 = - y^2 / 2 so

x^2 + y^2 = c

where c is the constant of integration.

Now, the condition is that y = 2 when x = 1 so:

(1)^2 + (2)^2 = 1 + 4 = 5

so the particular solution is: x^2 + y^2 = 5

(Aside question: Can readers figure out why the fractional form x^2 /2+ y^2/2 was discarded?)

Now let's examine a more complex 1st order equation more worthy of the brainpower of the typical readers of this blog:

Find the general solution and the particular curve passing through the point (0,0) of the differential equation:

exp(x) cos(y) + (1 + exp(x)) sin(y) dy = 0

This looks a bit fearsome, but again, the first rule is simplify, which means separating variables (this is also usually where one's acumen with basic algebra comes in!)

we obtain:

exp(x)/ (1 + exp(x)) + [sin(y)/ cos(y)] dy

= exp(x)/ (1 + exp(x)) + tan(y)dy = 0

We then integrate this to obtain*:

ln(1 + e^x) - ln cos(y) = ln c

Or:

ln(1 + e^x) = ln c + ln cos(y)

where again, c is the constant of integration. We can easily simplify the above (using well known principles of natural logs) to get:

1 + e^x = c cos (y)

And this is the general solution.

To get the particular solution we need to substitute the ordered pair values for (0.0) into the general solution, whence:

1 + e^(0) = c(cos (0))

so that: 1 + 1 = c

and c = 2

then we get: 1 + e^x = 2 cos (y)


Now for some problems:

1) Show that y = cx^2 - x is a solution of the DE: xy' = 2y + x

2) Show that y = (2x + c) exp(-x) is a solution of the DE: dy/dx + y = 2 exp(-x)

3) For the differential equation: dy/dx = -x/4y

sketch the curve which passes through the point (1,1)

4) Find the general solution of: sin (t) dp + p cos(t) dt = 0

5) Find the particular solution for the equation:

exp(x) sec(y)dx + (1 + e^x) sec (y) tan(y) dy = 0

and y = 60 deg when x = 3

* (Tables of integrals are often also used to integrate and solve DEs!)

Solutions to the odd ones next time!