Differential Equations Homework

[ballajr]

Homework Statement

1. Let D and I be the differential operator and the identity operator, respectively. Find two
real-valued functions f(t) and g(t) such that:
D^{2} + I = (D + f(t)I)(D + g(t)I):

2. Use this theorem to prove the corollary given below.
Theorem: There are two nonzero solutions y_{1} and y_{2} 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 c_{2}y_{1}+c_{2}y_{2} for arbitrary constants c_{1} and c_{2} is a general solution to the
differential equation.
Corollary: If z_{1} and z_{2} are two nonzero solutions to the differential equation such that one of the two functions is not a constant multiple of the other, then c_{1}z_{1} + c_{2}z_{2} for arbitrary constants c_{1} and c_{2} is a general solution the
differential equation.

3. Prove the theorem stated in #2.

Homework Equations

1. I know there is a theorem such that ((D^{2} + a^{2}I) = 0 and then this equation equals c_{1}cos(at) + c_{2}sin(at)

The Attempt at a Solution

1. I tried to do this, but got lost. I know that
(D^{2} + I) = c_{1}cos(1t) + c_{2}sin(1t).
That's all I've got.

2. My teacher assigned this problem, but I have no clue how to go about proving this corollary. The corollary seems like the same as the theorem except with using z's instead of y's and I'm confused.

3. Again, I have no clue how to prove this theorem, so I need some help.


Thanks guys!

 

[ballajr]

For some reason, it didn't subscript the c's and y's and z's. All of those are supposed to be written as subscripts, not superscripts. The only superscripts are just the D's and they are squared.

 

[HallsofIvy]

Any thing inside the [ tex ] tags tends to get put above the line anyway.

Put the whole thing inside [ tex ]: [ tex ]c_2[ /tex ]: c_2, not c[ tex ]_2[ /tex]: c_2.

In fact, it is better to put the whole formula inside the tex tags.


This problem has nothing to do with solving the equation.

You are asked to find two functions, f(x) and g(x) so that
(D+ f(x)I)(D+ g(x)I)= D^2+ I

Go ahead and "multiply" the left side: (D+ f(x)I)(D+ g(x)I)= D(D+ g(x)I)+ f(x)I(D+ g(x)I)
= D^2+ D(g(x))I+ f(x)D+ f(x)g(x)I= D^2+ g'(x)I+ f(x)D+ f(x)g(x)I.

For that to be the same as D^2+ I, you must have f(x)= 0 and g'(x)+ f(x)g(x)= 1. The first of those equations makes it particularly easy!

 

[ballajr]

Thank you so much! That seems so easy, but I don't think I would have ever known that method, but now it makes so much sense. Thank you!

Any input on the other 2 problems?

 

[ballajr]

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?