1  Why two terms?


Here's I'll explore why a second order linear homogeneous constant coefficient ordinary differential equation must always have two independent solutions. Actually, the proof I give below will work for linear ODE's whose coefficients are not constants also. Let's start with the differential equation

a d2y
dx2
 + b dy
dx
 + c y = 0.
Suppose we've found one solution, say y = y1(x). So, ay1"+by1�+cy1=0, since it's a solution. Suppose there is another solution, y2(x). Whatever this other solution is, I can write y2(x) = f(x)y1(x) for some function f(x). It's very simple. Just let
f(x) = y2(x)
y1(x)
 .
Since I don't know y2(x), I don't know f(x) either. However, I do know that y2 (just like y1) is a solution to the differential equation, so
a d2
dx
 2

 
 (f(x)y1(x)) + b d
dx
 (f(x)y1(x)) + c(f(x)y1(x))=0.
By the product rule,
d
dx
 (f(x)y1(x)) = f�(x)y1(x) + f(x)y1�(x)
and
d2
dxx
 (f(x)y1(x)) = f"(x)y1(x) + 2f�(x)y1�(x) + f(x)y1"(x).
If we substitute these into the differential equation, we get
a[f"(x)y1(x) + 2f�(x)y1�(x) + f(x)y1"(x)]+b[f�(x)y1(x) + f(x)y1�(x)]+cf(x)y1(x)=0.
I can rearrange this so that all the f(x) terms are together, and all the f�(x) terms, and the f"(x) term, giving
ay1(x)f"(x) + [2ay1�(x)+by1(x)]f�(x) +[a y1"(x) + by1�(x)+cy1(x)]f(x)=0.
Since ay1"(x)+by1�(x)+cy1(x)=0, the last term disappears, giving
ay1(x)f"(x) + [2ay1�(x)+by1(x)]f�(x)=0.
Now remember that y1(x) is known. Letting g(x)=f�(x), so f"(x)=g�(x), this is just a first order separable ordinary differential equation for g(x).
ay1(x) dg
dx
 + [2ay1�(x)+by1(x)]g=0, so

ay1(x) dg
dx
 = - [2ay1�(x)+by1(x)]g, so

a �
� 		dg
g
 = - �
� 		[2ay1�(x)+by1(x)]dx
y1(x)
 ,

= - 2a �
� 		y1�(x)dx
y1(x)
 - �
� 		bdx,

= - 2a lny1(x) - bx + C
Therefore (skipping a few steps),
g(x) = A e-(b/a)x
y1(x)2
Since g(x) is not the zero function, f(x) is not a constant function, so y2(x)=f(x)y1(x) is a distinct solution independent of y1(x).

2  Finding extra solutions?

By the way, this method can be used to explain why, when the roots of the auxiliary equation are equal (say k and k), the general solution is y=Axekx + Bekx.
If the auxiliary equation is (m-k)2, then differential equation is ay"+by�+cy=0, where a=1, b=-2k and c=k2. We already have y1(x)=ekx as a solution. Then, the formula above gives
g(x) = A e2kx
(ekx)2
 = A.
If g(x)=A, then f(x) = Ax+B, so y2(x) = f(x)y1(x) = (Ax+B)ekx.
You knew there had to be a real reason, didn't you?


File translated from TEX by TTH, version 3.76.
On 03 Oct 2006, 15:42.