Here is a short one about two simple differential equations. They both have “standard” solutions that appear in textbooks, but here is a method that treats them in similar ways.

First, an easy one. If is a (infinitely) differentiable function such that , then for some constant . Proof: Write . Then and hence is a constant function .

Here is another one. Let be a (infinitely) differentiable function defined on [ such that . Then for some constants and . First make the substitution , and set . Consider also the function . Then

and

.

Now we do the same for :

and

.

The numerator of the last term is actually a constant: its derivative is

,

which is 0 by the fact that .

Hence for some constant . Integrating twice, we get for some constants and . Now substituting back , we get . Since , we conclude that .

-Steven

for the second problem ,

If m is a positive real number , the only solution to differential equations :

f”(x) + mf(x) = 0

is of the form :

f(x) = a cos(nx) + b sin(nx) , where n = (m)^1/2

Consider the function :

g(x) = f(x)cos(nx) – (1/n)*f'(x)sin(nx)

h(x) = f(x)sin(nx) + (1/n)*f'(x)cos(nx)

By differentiating g(x) and h(x) , we get :

g'(x) = 0 = h'(x)

so they’re just constant function.Hence , there exist a , b real numbers such that :

a = g(x) = f(x)cos(nx) – (1/n)*f'(x)sin(nx)

b = h(x) = f(x)sin(nx) + (1/n)*f'(x)cos(nx)

and we get :

f(x) = acos(nx) + bsin(nx)

Q.E.D.

By:

axison September 23, 2010at 4:14 PM