4

Given the function:

$x+ye^{-x} \frac {dy}{dx}=0$ where $y(0)=1$ and you were told to solve it, I know I must multiply by $e^x$ giving $xe^x+y \frac {dy}{dx}=0$. Then re-arranging to give $ydy=-xe^xdx$.

After this I proceed to integrate both sides:

$\int ydy = -\int xe^xdx$

My professor uses the notation $\int_{}^{y} ydy = \int_{}^{x}xe^xdx$ which makes complete sense to me, however, she insists that we must insert "dummy variables" into where the limits $x$ and $y$ are giving:

$\int_{}^{y} udu = \int_{}^{x}te^tdt$

I am fully willing to use this from now on but I can't seem to understand/see why we must replace $x$ and $y$ with $t$ and $u$ respectively. I remember it being to do with the fact that if you didn't do this you can equate $arcsin$ and $arctan$ or something similar. Could someone give me some guidance, much appreciated.

Lo-urc
  • 567
  • 2
    The basic idea is that you can't use the same symbol for two different things. In $\int^y y,dy$, $y$ is both a variable and a limit of integration. – Matthew Leingang Nov 21 '17 at 19:49
  • 2
    Do you know anything about programming? The variable of integration is in what programmers call a "local scope", which means that technically it isn't seen on the outside of the integral and can be re-used outside the integral. But abusing local scope by using the same variable just a little bit outside the scope boundary as well as within the scope boundary is confusing for humans. – Ian Nov 21 '17 at 19:49
  • 1
    Also just in general, math education should be trying to move you away from the idea of computing antiderivatives and toward the idea of computing definite integrals with possibly variable limits, which you just can't do when the symbol is shared. – Ian Nov 21 '17 at 19:49
  • 1
    The limits in the integral tells you what values of the integrating variable are relevant. You can't talk about letting $y$ be between $0$ and $y$. – Arthur Nov 21 '17 at 19:51
  • Thank you very much all, makes complete logical sense now. – Lo-urc Nov 21 '17 at 20:00
  • Not addressing your question, but in your derivation you go from $$x+y \frac {dy}{dx}=0$$ to $$xe^x+y \frac {dy}{dx}=0$$which is incorrect. The correct equation would be $$xe^x+ye^x \frac {dy}{dx}=0$$ You have to multiply the entire equation (both sides and distribute to each term) to come up with another true equation. One term just doesn't work. – Paul Sinclair Nov 22 '17 at 02:10
  • Yes I made a mistake in writing my initial equation, I'll change it soon. – Lo-urc Nov 22 '17 at 09:50

1 Answers1

2

Here's a somewhat visual way to think about it: $ \int_{a}^{b} f(t) dt $ represents the signed area between the t-axis and the function f(t) over the interval $a < t < b$. If we allow b to change according to a variable (say $x$), our integral becomes a function of $x$: $$F(x) = \int_{a}^{b(x)} f(t) dt $$ Our new function, $F(x)$, would visually represent what would happen if you integrated with different bounds.

Can b be a function of t instead? No. $F(t) = \int_{a}^{b(t)} f(t) dt $ would represent the signed area between the t-axis and the function f(t) over the interval $a < t < b(t)$. As t would change, $t$ wouldn't necessarily (in general) be in the interval $(a, b(t))$. It just wouldn't make sense.

Thomas Davis
  • 1,273