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I saw this as a step in a calculation and it was confusing. The left cancels down to:

$$\int \frac{d^2y}{dx^2}dy$$

But surely the answer is $\frac{d^2y}{dx^2}y$; instead, $\frac {d^2}{dx^2}$ is being treated as a constant and not $\frac {d^2y}{dx^2} $

https://www.wolframalpha.com/input/?i=∫%28d%5E2y%2Fdx%5E2%29+dy. Putting in the calculation directly yields the result I would have expected.

user577215664
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Lewis Kelsey
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1 Answers1

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For the left side you have that: $$I=\int y''y'dx= \frac 12\int(y'^2)'dx=\frac 12 \int \dfrac {d(y')^2}{dx}dx$$ $$I=\frac 12 \int d(y')^2=\frac 12 (y')^2$$

Note that; $$(y'^2)'=2y'y''$$

user577215664
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    so why do you get a different answer if you cancel and integrate for dy? @LewisKelsey Because from what you wrote $$ \int y''dy=\int \dfrac {dy'}{dx}dy=\int y' dy'=\frac 12 y'^2+c$$ It gives the same answer of course. – user577215664 Jun 24 '20 at 01:37
  • I couldn't get Wolfram alpha to give any answer other than y''y for that but I'm not surprised it can't be done like that. – Lewis Kelsey Jun 24 '20 at 10:35
  • I see. WA is a good soft but sometimes it may fail. Yeah it can be done that way too very simple. @LewisKelsey Your integral is just $I=\int y'dy'$ – user577215664 Jun 24 '20 at 11:08
  • as for another question, $\int \frac{dy} {dx} dy $ which WA gave an incorrect answer on there appears to be no answer. I ended up at $\int \left(\frac{dy}{dx}\right) ^2 dx$ – Lewis Kelsey Jun 24 '20 at 13:08
  • Yes sometimes you cant evaluate the integral as in this case @LewisKelsey – user577215664 Jun 24 '20 at 13:09
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    @Aryadeva: Why does $\int \frac{dy'}{dx}dy=\int y'dy'$? I think this is because you moved the $dx$ under the $dy$ but I'm not sure if you can do this. I might ask a new question on Math SE about this. – Axion004 Jun 24 '20 at 13:49
  • @Axion004 yes thats what i considered. but you can also change do this by variable change – user577215664 Jun 24 '20 at 14:26
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    You have that $dy=y'dx$ I put this in the integral then I have $$I=\int y'dy'$$ @Axion004 But it will be great to open such a topic with calculus tag. – user577215664 Jun 24 '20 at 14:40
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    @Axion004 Basically it goes like this $$\int \frac{d^2y}{dx^2} \frac{dy}{dx} dx \equiv \int \frac{d^2y}{dx^2}{dy}$$ $$\equiv \int \frac{d\frac{dy}{dx}}{dx} dy$$ $$\equiv \int d\frac{dy}{dx} \frac{dy}{dx}$$ $$\equiv \int dy'y'$$ $$\equiv \int y'dy'$$ $$\equiv \frac{1}{2} y'^2$$ it looks alright to me – Lewis Kelsey Jun 25 '20 at 16:37
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    wolfram alpha will produce 1/2y'^2 for the former and y''y for the latter, which is what confused me – Lewis Kelsey Jun 25 '20 at 16:47
  • @LewisKelsey Axion has posted on this topic maybe it's of some intrest to read. – user577215664 Jun 25 '20 at 16:54
  • @LewisKelsey take a look here https://math.stackexchange.com/questions/3733044/when-can-you-divide-out-dx-in-an-integral-as-if-it-is-a-fraction – user577215664 Jun 25 '20 at 16:56