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Consider the following exercise:

Using the Laplace transform, find a solution $y(x)$ of the following initial value problem:

$$\begin{cases} y'' +y = x + 1, \quad x > \pi \\ y(\pi) = \pi^2 \\ y'(\pi) = 2\pi \end{cases}$$

Suggestion: Make the change of variables $t = x - \pi$.

Of course, this equation is pretty easy to solve without the Laplace transform, but the whole point of the exercise is to use it. That's not the problem, though. I'm having some trouble with the change of variables.

I know that for the derivatives you have to use the chain rule; in this case, since $dt = dx$ there's no difference. My main issue is with the initial conditions. I'm not very sure how to restate them in terms of $t$, and I think that's because I don't really know a precise definition of change of variables. How is it done? Do we define a new function, something like $g(t) = y(x-\pi)$, if that even makes sense? What would be a general method to make sure one does things like this carefully?

I apologize if the question is rather vague. Just to be clear, I'm not asking how to solve this particular differential equation; I've realized I get confused in general when using change of variables in an ODE.

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Try to use Laplace transform to solve: $$\begin{cases} g'' +g = t+ \pi + 1, \quad t > 0 ,\\ g(0) = \pi^2, \\ g'(0) = 2\pi. \end{cases}$$

Like you said, but it is not letting $g(t) = y(x-\pi)$, rather it is letting $g(t) := y(t+\pi) = y(x)$. Basically the equation is translated by $\pi$ in the independent variable, so when originally the initial condition is at $x = \pi$, now it is at $t = x-\pi = 0$.

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That works. But what's the point then ofusing two different – Javier Jun 30 '13 at 13:09
@JavierBadia Because Laplace transform's formula for derivatives involve the initial value at 0: $\mathcal{L}\{f'\} = s\mathcal{L}\{f \}-f(0)$. You can use the initial value at $\pi$ as well, but whenever you perform the integration by parts using the definition of Laplace transform, you will find it essentially you are translating the coordinates as well, so it is easier to do it before going into the Laplace transform stage. – Shuhao Cao Jun 30 '13 at 16:03
@Shuhaho: I'm sorry, I accidentally submitted that comment early and was going to fix it but I forgot. I understand that. What I was going to say is, what's the point of using a different variable name? Like you said; isn't it clearer to say "use $g(x) = g(x+\pi)$" instead of "use $t = x - \pi$"? – Javier Jun 30 '13 at 16:43
@JavierBadia Using the same letter would cause confusion, for $g(x) = g(x+\pi)$ means periodicity for the same function $g$, using a different letter just avoid this: say $y(x) = x^2$, then letting $g(t) := y(t+\pi) = t^2 + 2\pi t + \pi^2$, this can be also written as $g(x) := y(x+\pi) = x^2 + 2\pi x + \pi^2$, $g$ and $y$ are simply different functions, one of which is a translation in $x$ of the other. – Shuhao Cao Jun 30 '13 at 16:58
It seems I just can't write comments today. I meant $g(x) = y(x+\pi)$. – Javier Jun 30 '13 at 17:04

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