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Suppose $y_1$ is a solution of a non-homogeneous linear D.E. and $y_2$ is a solution to a homogeneous equation. Which of the following is/are true?

a. $-y_1$ is a solution to the non-homogeneous equation.

b. $-y_2$ is a solution to the homogeneous equation.

c. $y_1-y_2$ is a solution of the non-homogeneous equation.

d. $y_2-y_1$ is a solution of the homogeneous equation.

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You should shed some light on how the non-homogenous and the homogenous linear DE are connected or look like, pick an example otherwise i find your question some what ill formed. There has to be no reason for b and c to hold true if the equations look completely different also you might wanna check out the theorems about uniqueness and existence of solutions of non-homogenous linear DE. –  Hardy Apr 3 '12 at 21:10
    
I did. Nothing is really making sense. He said something about "plugging it in" and seeing what it comes out to, but we were running out of time so I didn't catch it. And there is no other information. That's pretty much it. –  Nico Bellic Apr 3 '12 at 21:18
    
@Hardy: As usual in these problems, the equation is the same for both $y_1$ and $y_2$, only the homogeneous version has the RHS of $Ly=b$ cut off to make $Ly=0$. Uniqueness/existence considerations are not relevant to this problem. –  anon Apr 3 '12 at 21:35
    
@anon I totally agree with the comments you made about the equation, but none the less it would be nicer and clearer if the OP mentioned that. My sole intention behind the comment, was to bring it to the OP's knowledge. I could have been misquoting the name of the result i wanted to bring up as i was in hurry to leave for work. My apologies if that wasted anyone's time. –  Hardy Apr 3 '12 at 23:10
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1 Answer

up vote 3 down vote accepted

The homogoneous DE is $Ly=0$, where $L$ is a linear differential operator, such as (for example)

$$Ly:= y\,''-x^2y.$$

Observe that for the above operator, if $u,v$ are functions and $\alpha,\beta$ constants, we have

$$L(\alpha u+\beta v)=\alpha Lu+\beta Lv.$$

This is what it means to be linear. We are given the following two facts ($b$ is a function):

$$Ly_1=b, \quad Ly_2=0.$$

Note $b$ is not the zero function and $L$ is now arbitrary. Using these, the four questions become:

  1. Does $L(-y_1)=b$?
  2. Does $L(-y_2)=0$?
  3. Does $L(y_1-y_2)=b$?
  4. Does $L(y_2-y_1)=0$?

Using the linearity properties of $L$, reduce the above expressions to expressions in just $Ly_1$ or $Ly_2$, then replace these respectively with $b$ and $0$, and then decide if the resulting equation is true/false.

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My professor did mention something about "plugging back in" but I didn't catch exactly what. I was frantically looking for some explanation in my 'existence and uniqueness' section of the book but I didn't even think to look for linearity. Thanks a lot, this was a pretty easy question now that you explained it. –  Nico Bellic Apr 3 '12 at 21:23
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