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Hello I just have a short question about a remark made in my first class of intro ODE. My Professor was just motivating with a simple example, he wrote, $$\frac{dy}{dx}=y(x)$$

So of course it was clear that he was referring in general to $y(x)=ce^{x}$ where $c \in \mathbb{R}$

and then he also added and for all $ \ x \in \mathbb{R}$.

Now here is where I got briefly confused , because I thought for example consider $y(x)=e^9$, then $\frac{dy}{dx}=0 \neq e^{9}$

So what I took this to mean is that $y(x)=ce^x$ is the solution, and if you evaluate it at any x the equation is true, i.e. simply because $e^x$ was the solution? I just want to make sure that is what is meant in this context, or if there was some error in my understanding/notation?

Thanks all.

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$e^9$ is a constant, and so is not of the form $ce^x$.

It is true that $y = c e^x$ is a solution to the differential equation, where $c \in \mathbb R$ is any constant.

But this is a function of $x$ and so cannot give a derivative of zero unless it is a constant function.

It is analogous to saying:

Let $z(t) = t^2$ for $t \in \mathbb R$. Then $z(t) = 9^2$.

This is false; you can say $z(9) = 9^2$, for example, but not $z(t) = 9^2$.

This might help:

Consider $y_1 = e^x$. Then $y_1(9) = e^9$, and also $\left.{\dfrac {\mathrm dy_1}{\mathrm dx}}\right\vert_{x = 9} = e^9$.

This relationship between $y_1$ and $\dfrac {\mathrm dy_1}{\mathrm dx}$ would hold for all $x \in \mathbb R$.

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  • $\begingroup$ That is what I am asking, because if you read it I wrote it was said x in R $\endgroup$ – Quality May 5 '15 at 2:05
  • $\begingroup$ You did not write that $x \in \mathbb R$, but I will try to clarify my answer in any event. $\endgroup$ – GFauxPas May 5 '15 at 2:06
  • $\begingroup$ Yes I did.. please look $\endgroup$ – Quality May 5 '15 at 2:07
  • $\begingroup$ Oh, I see. Wait one moment. $\endgroup$ – GFauxPas May 5 '15 at 2:08
  • $\begingroup$ Okay great thanks for the edit, so that is what is meant by a solution for all x $\endgroup$ – Quality May 5 '15 at 2:14
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There is a difference between $y(x)=ce^{x}$ and $y(x)=e^{9}$. The first one is an exponential function, but the second one is a horizontal line $y(x)=e^{9}=8103.08$. The derivative of the first one follow the rule the second one is just a constant number and will give a derivate of 0.

When one says x is defined as a real number it does not mean that you can just replace it with a number like this $y(x)=ce^{9}$. It means you can replace the input value, let's say $y(9)$ and solve for $y(x)$ at x=9.

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  • $\begingroup$ Yes so why would the professor write e^x for all x in R? Was it a mistake $\endgroup$ – Quality May 5 '15 at 2:06
  • $\begingroup$ No, if you write it like this $y(x)=e^{9}$ it's different from $y(x)=e^{x}$, in the first one there is no x. $\endgroup$ – Continuum May 5 '15 at 2:11
  • $\begingroup$ Yes I understand that, but by writing for ALL x this implies we could take x=9 $\endgroup$ – Quality May 5 '15 at 2:12
  • $\begingroup$ @Quality The solution to that equation (taking $y(0)=1$) is indeed equal to $e^9$ at $x=9$, but it is not also equal to $e^9$ at, say, $x=8$, where it is instead equal to $e^8$. $\endgroup$ – Ian May 5 '15 at 2:13
  • $\begingroup$ @Ian Thanks, I think that is more where I am confused. $\endgroup$ – Quality May 5 '15 at 2:14
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The formulation

"The solution is $y(x)=ce^x$ for some $c$ and all $x\in \Bbb R$"

is short and casual for the more correct statement that

"the solution of this ODE is a differentiable function $y:\Bbb R\to\Bbb R$ with function values $y(x)=ce^x$ (for all $x\in \Bbb R$)".

You will need to remember that a differential equation is a functional equation, i.e., its solutions are always functions. Sometimes, and often for textbook exercises, you can even give a computational algorithm for the values of this function.

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