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I hope that's not an extremely stupid question, but it' been in my mind since I was taught how to solve differential equations in secondary school, and I've never been able to find an answer.

For example, take the simple differential equation $$y'=y.$$

While I understand that $y=C\cdot e^x$ satisfies this equation, I do not understand why there can't be other solutions as well. So, how do we know this?

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    $\begingroup$ Picard's Theorem for a start... for linear odes I think you want to look at dimension arguments. $\endgroup$
    – JP McCarthy
    Feb 7, 2015 at 13:12
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    $\begingroup$ This is absolutely not a stupid question: in particular is not trivial for non-linear equations. $\endgroup$
    – MattAllegro
    Feb 7, 2015 at 13:17
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    $\begingroup$ For the special case $y'=y$: math.stackexchange.com/q/58097/1242 $\endgroup$
    – Hans Lundmark
    Feb 7, 2015 at 14:56
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    $\begingroup$ That's a very good question, though in many cases difficult to answer. And then, while studying more involved cases, one might discover that standard solutions are not enough (e.g. see weak solution) and matters complicate even more... $\endgroup$
    – dtldarek
    Feb 8, 2015 at 1:05
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    $\begingroup$ @MaxD To repeat what has been said, it's not stupid at all. Even more so when most of the time when solving a differential equation people just find some solutions and ignore the hard part which is to prove that those are the only solutions. In some cases, like the one you present, you don't need anything fancy to find all solutions. For instance this answer finds all solutions to a class of ODE which contains $y'=y$. $\endgroup$
    – Git Gud
    Feb 8, 2015 at 13:02

3 Answers 3

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When you study the topic in University, you will probably encounter uniqueness theorems that show that for certain ODE's, not only does a solution exist, but this solution is also unique - there are no other solutions.

For first order ODE's such as that in your example, the relevant theorem is the Picard–Lindelöf theorem, which shows that your solution is indeed unique around a given interval. These intervals can then be strung together to show this is the only solution across the entire domain.

I'll add that for some other forms of differential equations it is not at all clear that solutions are unique. The most famous example is probably the Navier-Stokes equations in three dimensions, which govern the motion of gases and liquids all around us, and yet no proof of uniqueness in all cases is known.

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    $\begingroup$ Thanks! I'm afraid the Picard-Lindelöf theorem is way to complicated for me to ever understand it, but at least your answer shows me that my question makes some sense. $\endgroup$
    – MaxD
    Feb 7, 2015 at 13:48
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    $\begingroup$ @MaxD - your question makes a lot of sense - it's occupied and still occupies the minds of many great mathematicians. $\endgroup$
    – nbubis
    Feb 7, 2015 at 14:05
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    $\begingroup$ @MaxD: That theorem applies to a general class of differential equations, which is why it is complicated. For linear ordinary differential equations like the one you mentioned, knowing the form of the solution means that we can do a similar trick as I did in my answer to prove that it is unique. It of course fails when we do not know the form of the solution. $\endgroup$
    – user21820
    Feb 7, 2015 at 14:32
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For any solution $y$ to that equation, the following holds.

$y e^{-x} - y'(x) e^{-x} = (y(x)-y'(x))e^{-x} = 0$ for any $x \in \mathbb{R}$.

Thus $\frac{d}{dx}(y(x)e^{-x}) = 0$ for any $x \in \mathbb{R}$.

Thus $y(x)e^{-x} = c$ for any $x \in \mathbb{R}$, for some constant $c \in \mathbb{R}$ [by mean value theorem].

Note that this includes even the case of $c = 0$, and also does not face division by zero, unlike the typical method taught in high-school.

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  • $\begingroup$ Sir I would like to know, how & where ,the mean value theorem is used here? $\endgroup$
    – Akash Patalwanshi
    Sep 5, 2018 at 3:10
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    $\begingroup$ @AkashPatalwanshi: The general claim is that if a real function $f$ has zero derivative on some interval $I$, then $f$ is constant on $I$. Proof: If $f$ is not constant on $I$, then for some points $x,y$ in $I$ we have $f(x) ≠ f(y)$, but since $f$ is differentiable on $[x,y]$, by the mean value theorem there must be some point $z$ in $[x,y]$ such that... Can you finish from here? =) $\endgroup$
    – user21820
    Sep 5, 2018 at 3:19
  • $\begingroup$ Thank you so much sir :-) further I would like to know, why you started by considering $ye^{-x}-y'e^{-x}$ ? $\endgroup$
    – Akash Patalwanshi
    Sep 5, 2018 at 3:22
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    $\begingroup$ @AkashPatalwanshi: Because it works! Sometimes there are things like this where there is no reason other than "it works". More generally, if you have the differential equation $\frac{dy}{dx}+f(x)·y = 0$ you can multiply both sides by $\exp(\int f(x)\ dx)$ and then observe that the left-hand side is equal to $\frac{d}{dx}(y·\exp(\int f(x)\ dx))$ by the product and chain rule. And then just like here you conclude that $y·\exp(\int f(x)\ dx)$ is constant. Similarly for $\frac{dy}{dx}+f(x)·y = g(x)$. It 'just works' because $\exp' = \exp$. This is also called the integrating factor method. $\endgroup$
    – user21820
    Sep 5, 2018 at 3:32
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    $\begingroup$ @AkashPatalwanshi: You're welcome! =) By the way, you can usually find me in the Logic chat-room or the Basic Math chat-room. $\endgroup$
    – user21820
    Sep 6, 2018 at 6:54
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You expressed a special case. I respond to you by that case:

If $$y'=y$$

it means that

$$\frac{y'}{y}-1=0$$

Then we integrate of both sides:

$$ln(y)-x=C$$

Can integration of zero have any result other than a constant number? If cannot, then that include the whole answers.

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    $\begingroup$ This is an incomplete solution as it fails to handle the case of $y$ such that $y(x) = 0$ for some $x$. $\endgroup$
    – user21820
    Feb 7, 2015 at 13:20
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    $\begingroup$ @user21820 $y$ cannot be zero unless it is zero everywhere. which is covered by the answer $C_2 exp(x)$. I did not refer to that to avoid distraction. thanks for mentioning. $\endgroup$
    – Arashium
    Feb 7, 2015 at 13:22
  • $\begingroup$ Sure, the reason I mention it is that most high-school students are not taught to understand the methods used and as a result will fail to solve things like $xy' = 2y$. $\endgroup$
    – user21820
    Feb 7, 2015 at 13:26
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    $\begingroup$ Also, using logarithms glosses over the issue of branch cuts with negative $y$, which high-schools again usually ignore and stuff under the carpet by throwing absolute value symbols around, and then utterly mysteriously delete them when getting the solution. $\endgroup$
    – user21820
    Feb 7, 2015 at 13:28
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    $\begingroup$ Ah okay, well in the real world we generally get infinitely differentiable functions and so lots of special cases just disappear. But I do suspect the asker was really looking for real mathematical justification. $\endgroup$
    – user21820
    Feb 7, 2015 at 13:37

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