

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 
d^{2}y
dx^{2}

+ b 
dy
dx

+ c y = 0. 

Suppose we've found one solution, say y = y_{1}(x). So, ay_{1}"+by_{1}�+cy_{1}=0, since it's a solution. Suppose there is another
solution, y_{2}(x). Whatever this other solution is, I can write y_{2}(x) = f(x)y_{1}(x) for some function f(x). It's very simple. Just let
f(x) = 
y_{2}(x)
y_{1}(x)

. 

Since I don't know y_{2}(x), I don't know f(x) either. However, I do know that y_{2} (just like y_{1}) is a solution to the differential
equation, so
a 
d^{2}
dx

2

(f(x)y_{1}(x)) + b 
d
dx

(f(x)y_{1}(x)) + c(f(x)y_{1}(x))=0. 

By the product rule,

d
dx

(f(x)y_{1}(x)) = f�(x)y_{1}(x) + f(x)y_{1}�(x) 

and

d^{2}
dx^{x}

(f(x)y_{1}(x)) = f"(x)y_{1}(x) + 2f�(x)y_{1}�(x) + f(x)y_{1}"(x). 

If we substitute these into the differential equation, we get
a[f"(x)y_{1}(x) + 2f�(x)y_{1}�(x) + f(x)y_{1}"(x)]+b[f�(x)y_{1}(x) + f(x)y_{1}�(x)]+cf(x)y_{1}(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
ay_{1}(x)f"(x) + [2ay_{1}�(x)+by_{1}(x)]f�(x) +[a y_{1}"(x) + by_{1}�(x)+cy_{1}(x)]f(x)=0. 

Since ay_{1}"(x)+by_{1}�(x)+cy_{1}(x)=0, the last term disappears, giving
ay_{1}(x)f"(x) + [2ay_{1}�(x)+by_{1}(x)]f�(x)=0. 

Now remember that y_{1}(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).
ay_{1}(x) 
dg
dx

+ [2ay_{1}�(x)+by_{1}(x)]g=0, so 

ay_{1}(x) 
dg
dx

=  [2ay_{1}�(x)+by_{1}(x)]g, so 

a 
� �


dg
g

=  
� �


[2ay_{1}�(x)+by_{1}(x)]dx
y_{1}(x)

, 

=  2a 
� �


y_{1}�(x)dx
y_{1}(x)

 
� �

bdx, 

=  2a lny_{1}(x)  bx + C 

Therefore (skipping a few steps),
g(x) = 
A e^{(b/a)x}
y_{1}(x)^{2}



Since g(x) is not the zero function, f(x) is not a constant function, so y_{2}(x)=f(x)y_{1}(x) is a distinct solution
independent of y_{1}(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=Axe^{kx} + Be^{kx}.
If the auxiliary equation is (mk)^{2}, then differential equation is ay"+by�+cy=0, where a=1, b=2k and c=k^{2}.
We already have y_{1}(x)=e^{kx} as a solution. Then, the formula above gives
g(x) = 
A e^{2kx}
(e^{kx})^{2}

= A. 

If g(x)=A, then f(x) = Ax+B, so y_{2}(x) = f(x)y_{1}(x) = (Ax+B)e^{kx}.
You knew there had to be a real reason, didn't you?
File translated from
T_{E}X
by
T_{T}H,
version 3.76. On 03 Oct 2006, 15:42.
 