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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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    @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.2012-04-03

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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.2012-04-03