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I'm trying to understand the concept of Second Order ODE general solution, and I need help. Why does a quotient of basis of a general solution $\neq constant$?

For example: $y_1=e^x$ and $y_2=e^{-x}$ are solutions to ODE $y''-y=0$

Before stating general solution, it goes on to check that $$e^x/e^{-x}\neq constant$$ Why is that? And what is this constant? Do they mean to say that "x" don't cancel as in $2x/2x=1$, and?

Here is the passage from the book that I'm referring to: General Solution to second order ODE

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  • $\begingroup$ Improvements or edits are welcome. $\endgroup$
    – Jek Denys
    Commented Jul 21, 2016 at 23:43
  • $\begingroup$ Why do you expect the quotient of some basis elements to be a constant? $\endgroup$
    – Andrew Whelan
    Commented Jul 21, 2016 at 23:46
  • $\begingroup$ Well, that's what I'm confused. That's what it says in the book. $\endgroup$
    – Jek Denys
    Commented Jul 21, 2016 at 23:46

2 Answers 2

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If the quotient of two solutions is constant that means that one is a constant multiple of another, which means that one is linearly dependent on the other, which means they cannot form a basis of solutions (a basis is a linearly independent and spanning set of solutions).

For instance, $e^x$ and $2e^x$ are both solutions to your ODE, but clearly they are linearly dependent since they have a quotient of $2$, so they can't form a basis.

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  • $\begingroup$ So what would it mean when two solutions can't form a basis? can these solutions still be useful somehow? $\endgroup$
    – Jek Denys
    Commented Jul 22, 2016 at 0:05
  • $\begingroup$ The most general solution to a differential equation is given by an arbitrary linear combination of the solutions which form a basis. So here we have two linearly independent solutions ($e^x$ and $e^{-x}$) so the most general solution to your differential equation is given by $c_1 e^x + c_2 e^{-x}$. Your initial/boundary conditions determine the constants. $\endgroup$
    – Andrew Whelan
    Commented Jul 22, 2016 at 0:11
  • $\begingroup$ If two solutions can't form a basis then only one of them is really useful. For any given differential equation there are an infinity of solutions. The point of having a basis of solutions is so that you can characterise all the solutions given any initial or boundary conditions. $\endgroup$
    – Andrew Whelan
    Commented Jul 22, 2016 at 0:14
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    $\begingroup$ Thank you for clarifying the purpose of forming basis. Until you did I didn't even realize, that I had no idea what it was used for :D $\endgroup$
    – Jek Denys
    Commented Jul 22, 2016 at 12:13
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    $\begingroup$ No problem, glad to be of use! $\endgroup$
    – Andrew Whelan
    Commented Jul 22, 2016 at 12:41
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The general solution of a homogeneous linear ODE of order $n$ consists of the linear combinations of a basis consisting of $n$ linearly independent solutions. In the case $n=2$, for two solutions to be linearly independent is equivalent to neither being a constant multiple of the other, i.e. neither is identically $0$ anThe theory says what you need is a basis consisting of $n$ (in this case $2$) linearly independent solutions. d their quotient is not constant.

EDIT: Sorry to sound like a book, but that's just the way I am.

This all comes from linear algebra, which it would be handy to know.

Let's stick to the case $n=2$. Say you've found two solutions $f_1(x)$ and $f_2(x)$. It would be very bad if $f_2$ was a constant multiple of $f_1$, say $f_2(x) = c f_1(x)$, because then linear combinations of $f_1$ and $f_2$ doesn't give you any more solutions than you get with just $f_1$: $a f_1(x) + b f_2(x) = (a + b c) f_1(x)$. So you check that $f_2(x)$ is not a constant multiple of $f_1(x)$ by seeing that $f_2(x)/f_1(x)$ is not constant.

In this example, with $f_1(x) = e^{-x}$ and $f_2(x) = e^x$, $f_2(x)/f_1(x) = e^x/e^{-x} = e^{2x}$, and that is not constant. Or you could take $f_1(x)/f_2(x)$: again that isn't constant, so $f_1(x)$ is not a constant multiple of $f_2(x)$.

The theory tells you (and it's not hard to prove) that this is all you need to conclude that $f_1(x)$ and $f_2(x)$ are linearly independent and their linear combinations give you the general solution.

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  • $\begingroup$ sorry mate, but you sound like my book, and I don't get you :(. Individual terms make sense, but I can't paint a big picture still. $\endgroup$
    – Jek Denys
    Commented Jul 21, 2016 at 23:50
  • $\begingroup$ Thank you for the insightful edit with the examples! This helped! =D $\endgroup$
    – Jek Denys
    Commented Jul 22, 2016 at 0:28
  • $\begingroup$ If I were to find the prove of the theory, that you mention in the last paragraph, where would I look for it? $\endgroup$
    – Jek Denys
    Commented Jul 22, 2016 at 0:29
  • $\begingroup$ For example, you might look here. $\endgroup$
    – Robert Israel
    Commented Jul 22, 2016 at 1:30

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