# Math Help - Prove this Theorem and Corollary

1. ## Prove this Theorem and Corollary

1. Use this theorem to prove the corollary given below.

Theorem: There are two nonzero solutions y1 and y2 to the differential equation
y''+ p(t)y' + q(t) = 0
such that one of the two functions is not a constant multiple of the other, and
that c2y1+c2y2for arbitrary constants c1 and c2 is a general solution to the
differential equation.

Corollary
: If z1 and z2 are two nonzero solutions to the differential equation such that one of the two functions is not a constant multiple of the other, then c1z1 + c2z2 for arbitrary constants c1 and c2 is a general solution the
differential equation.

3. Prove the theorem stated in #2.

2. My teacher gave the following hint:
The theorem states the existence of such two functions, and we don't know what they are. What we know is that there ARE two such functions. That's what the theorem is about, namely, the existence of such two functions. Now, not knowing what those are, we pull out tricks and managed to find two solutions; for example, z1=t² and z2=t³. Using this example, I raise a question: are we going to reach all the solutions by considering
a t² + b t³
and all constants a and b?

3. The only difference between the theorem and the corollary is that the theorem states that [b]there exist[b] two functions which give the general solution and the corollary says that any two functions, which are not multiples of one another, will give the general solution.

Suppose we are given two functions z1 and z2, both of which satisfy the differential equation, neither of which is a multiple of the other. Further suppose y1 and y2 are the functions in the theorem. Since any solutions of the equation can be written as a "linear combination of y1 and y2, in particular, z1 and z2 can: z1(t)= Ay1(t)+ By2(t) and z2(t)= Cy1(t)+ Dy2(t).

Solve those equations for y1 and h2 as functions of z1 and z2. For example, if we multily the first equation by C we get Cz1= ACy1+ BCy2. If we multiply the second equation by A we get Az2= ACy1+ AD2.

Subtract to get Cz1- Az2= (BC- AD)y2 y2= C/(BC-AD) z1- A/(BC- AD) [b]as long as BC- AD is not 0. Can you show that "BC- AD= 0" is exactly the case when one of z1 and z2 are multiples of each other- and so if they are not, then BC- AD is not 0 and we can solve these equation.

Now, since C1y1+ C2y2 is the general solution, replace y1 and y2 by their expressions in terms z1 and z2.

As for "prove the theorem stated in #2", there is no "#2"!

4. Sorry, I meant #1. Prove the theorem.