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February 3rd, 2015, 06:30 PM   #1
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Basic Differential Equation Question

Hi everyone!
This is my first post on this forum and it's about a really basic concept in differential equations.
So I'm taking my first differential equations class in college. I had previously taken Calc 1, 2, and multivariable calculus. Our teacher does a great job of teaching techniques (how to solve separable, exact, first order linear, etc), but he doesn't really explain what it all means and how I can understand it.
Here's an example problem that I don't understand.

1). y'=2x My first question is what makes this question different than a normal calculus problem? Why can't we just integrate both sides and get y=x^2?
2). dy/dx=2x OK this makes sense because y'= dy/dx
3) dy=2xdx Algebra. This makes sense
4) ⌠dy=⌠2xdx So we're integrating both sides...
5) y=x^2+C This is what doesn't make sense to me. I thought that the integral of dy/dx was y. How can we take away the dx and still integrate to get y? Does dy=y'?

You differential equations experts are probably chuckling at my stupidity. But honestly, our teacher just dived into techniques without explaining what anything means. Hopefully someone can explain this before I get more confused as the course continues.

Thanks!

Last edited by skipjack; February 3rd, 2015 at 06:59 PM.
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February 3rd, 2015, 06:41 PM   #2
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Originally Posted by Mango pear View Post
Hi everyone!
This is my first post on this forum and it's about a really basic concept in differential equations.
So I'm taking my first differential equations class in college. I had previously taken Calc 1, 2, and multivariable calculus. Our teacher does a great job of teaching techniques (how to solve separable, exact, first order linear, etc), but he doesn't really explain what it all means and how I can understand it.
Here's an example problem that I don't understand.

1). y'=2x My first question is what makes this question different than a normal calculus problem? Why can't we just integrate both sides and get y=x^2?
2). dy/dx=2x OK this makes sense because y'= dy/dx
3) dy=2xdx Algebra. This makes sense
4) ⌠dy=⌠2xdx So we're integrating both sides...
5) y=x^2+C This is what doesn't make sense to me. I thought that the integral of dy/dx was y. How can we take away the dx and still integrate to get y? Does dy=y'?

You differential equations experts are probably chuckling at my stupidity. But honestly, our teacher just dived into techniques without explaining what anything means. Hopefully someone can explain this before I get more confused as the course continues.

Thanks!
What you see here is a problem on separation of variables. Let's see if rewriting a bit makes some more sense.

1) $\displaystyle y' = 2x$

2) $\displaystyle \frac{dy}{dx} = 2x$

Now let's do this a tad differently. I'm going to "multiply" both sides by dx and integrate:

3) $\displaystyle \int \frac{dy}{dx}~dx = \int 2x~dx$

4) Then $\displaystyle \int \frac{dy}{dx}~dx = y$ and $\displaystyle \int 2x~dx = x^2$

5) (Add in the constant of integration) $\displaystyle y = x^2 + C$.

Does that make it easier?

-Dan
Thanks from Mango pear

Last edited by skipjack; February 3rd, 2015 at 06:59 PM.
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February 3rd, 2015, 06:47 PM   #3
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Oh that makes more sense. So it's basically just explaining the intermediary steps that you skip over in normal calculus. Quick question: why do the dx's on the left not cancel out? Is it because we can make one dx a constant and put it on the outside of the integral?
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February 3rd, 2015, 07:12 PM   #4
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The dx specifies the variable with respect to which the integration is done.
$\displaystyle \int1dy$ and $\displaystyle \int dy$ are equivalent.
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February 3rd, 2015, 07:15 PM   #5
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Originally Posted by Mango pear View Post
1). y'=2x My first question is what makes this question different than a normal calculus problem? Why can't we just integrate both sides and get y=x^2?

5) y=x^2+C This is what doesn't make sense to me. I thought that the integral of dy/dx was y. How can we take away the dx and still integrate to get y? Does dy=y'?
1) The simple answer is that we can! However, this is (presumably) the first step into the world of differential equations which can and do (probably quite quickly in your course) get rather more difficult. Most differential equations cannot be solved simply by integrating, but it's good to learn the techniques on the easy ones.

5) Your step 3) is not, strictly speaking algebra. It is a notational shortcut that happens to work and make things easier. Topsquark's explanation is closer to the strict algebraic method, except that "multiplying by dx" is not an algebraic operation (as indicated by his quotes). In reality we are just integrating both sides of the equation with respect to x.

However, if we ignore the fact that step 3) is not really algebra, we got to $\int \mathrm d y = \int 1 \, \mathrm d y = y$. Note that here we are integrating with respect to $y$, so this is equivalent to the more normal $\int 1 \, \mathrm d x = x$ but with $y$s in place of $x$s. You will need to be familiar with the idea of integrating with respect to different variables in order to progress with differential equations.
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February 3rd, 2015, 08:05 PM   #6
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Originally Posted by Mango pear View Post
why do the dx's on the left not cancel out? Is it because we can make one dx a constant and put it on the outside of the integral?
${\mathrm d y \over \mathrm d x}$ is not a quotient, it is the differential operator ${\mathrm d \over \mathrm d x}$ operating on the function $y(x)$. Since there is no quotient in the expression $\int {\mathrm d y \over \mathrm d x} \, \mathrm d x$ the idea of cancelling the $\mathrm d x$s has no meaning.

Of course, in practice we sometimes do cancel them, because it works and makes the maths easier, but this is really just a notational shortcut - not algebra.
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February 3rd, 2015, 09:05 PM   #7
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OK I'm an idiot. I forgot that dx just tells whether you're going to take the derivative with respect to y or x. Ugh, I forgot Calc 1 stuff. Anyways, thanks for all of the help guys! One more question. In separable equations (for example here http://tutorial.math.lamar.edu/Class...Separable.aspx), if we're not actually "multiplying" dx, what is happening? Is there a way to rewrite a separable equation like topsquark rewrote my equation using the true notation? Sorry for all of the questions!

Last edited by Mango pear; February 3rd, 2015 at 09:41 PM.
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February 4th, 2015, 02:23 AM   #8
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$\displaystyle \int N(y)\frac{dy}{dx}dx = \int N(y)dy$

Note that a differential equation is sometimes written with $dy$ and $dx$ already separated, but that's just a way to emphasize that $y$ isn't necessarily to be thought of as the dependent variable.
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