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This is a lecture slide from one of my analysis lectures.

It uses the Mean Value Theorem to show that the only solutions to the particular differential equation are of the form given.

My question is, how does he know they are the only solutions? They are certainly solutions, but how has he proved that there is not another function that has this property?

Thank you. enter image description here

PhysicsMathsLove
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4 Answers4

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  1. In the proof above the following is shown: if $f$ is a solution of the Diff. Eq, then there is $A$ such that $f(x)=Ae^x$.

  2. On the other side it is easy to see that if $f$ is of the form$f(x)=Ae^x$, then $f$ is a solution of the Diff. Eq.

Fred
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In the proof, we did not assume anything on $f$ other than that $f$ is a solution (i.e., $f'(x)=f(x)$). And the discussion goes that if $f$ is a solution, then it must have the form $A e^x$ with some $A$ (the value $A$ is somewhat generic). Do I make sense?

shall.i.am
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  • I don't see where the must is shown. I appreciate that it is shown that it can have that form, but not that it has to be have that, and only that, form. – PhysicsMathsLove Jun 01 '17 at 12:03
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    @PhysicsMathsLove Do you see $f(x)=e^x g(x)$ holds? (This is how we defined $g$). Do you see $g$ must be a constant function? (For this, we used the corollary of yours and that $f$ is a solution whose form we do not know). – shall.i.am Jun 01 '17 at 12:22
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The theorem is a very standard one. And for uniqueness you do do need MVT. It is used in the last line where it says that $g $ is a constant based on $g'(x) =0$. This is a key step in the proof of the theorem which shows that ultimately any solution must be of the form $Ae^{x} $.

Another ingredient in the proof is mentioned in the beginning and that is the fact that $e^{x} $ is one solution. Note that this is not proved here, but rather is to be proved in the next chapter.

So based on existence of one solution and the use of MVT it is established that every solution must be of a specific form.

A more detailed analysis of the theorem is available in this answer.

Paramanand Singh
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The proof does not use the mean value theorem. It assumes a solution $f(x)$ and proves that $g(x)=e^{-x}f(x)$ must be a constant function, that is $g(x)\equiv A$, hence $f(x)=Ae^x$.

yoni
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  • It uses a corollary of the MVT. – PhysicsMathsLove Jun 01 '17 at 12:04
  • Yes. Still, to answer your question about uniqueness of form of solutions, we assume some solution and show it to be of the desired form. (With, as you stated, the help of a corollary of mvt) – yoni Jun 01 '17 at 12:09