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I'm a third year physics student, not a math student. The conventional way to solve this would be: $$\frac{d^2u}{dx^2}=\frac{d}{dx}(\frac{du}{dx})=1 \Rightarrow d(\frac{du}{dx})= dx\Rightarrow \int d(\frac{du}{dx})=\int dx\Rightarrow \frac{du}{dx}=x+A \Rightarrow du=(x+A)dx \Rightarrow u=\frac{x^2}{2}+Ax+B$$

So I'm wondering if what follows is correct: $$\frac{d^2u}{dx^2}=1 \Rightarrow d^2u=(dx)^2 \Rightarrow d(du)=dxdx \Rightarrow \int d(du)=\int dxdx \Rightarrow du=\int dxdx$$

At this point, I don't know if there are other ways to proceed, but I use integration by parts $$\int vdw=vw-\int wdv$$ by considering $v=dx$ and $dw=dx$, so that $w=x$ and $dv=d(dx)$: $$du=\int dxdx=xdx-\int x·d(dx)$$

If I assume that $-\int x·d(dx)=C$ is a constant, we get $$du=xdx+C \Rightarrow u=\frac{x^2}{2}+Cx+D$$ which is what we had originally. However, how would I go about proving that $\int x·d(dx)=constant$? Is it because $d(dx)=0$? And how do I prove that? Are there any other interesting ways to solve this differential equation?

elcocodrilotito
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Sorry, this is not correct. Already the first step $d^2 u = (dx)^2$ is meaningless.

Hans Lundmark
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  • Why, though? That was also a question I had – elcocodrilotito Nov 24 '18 at 17:07
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    @elcocodrilotito see https://math.stackexchange.com/questions/1784671/when-can-we-not-treat-differentials-as-fractions-and-when-is-it-perfectly-ok – shai horowitz Nov 24 '18 at 17:45
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    If I may counter with a question: why on earth would it be valid? What does it even mean? – Hans Lundmark Nov 24 '18 at 18:14
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    @shaihorowitz Thank you, that's probably where my confusion comes from – elcocodrilotito Nov 24 '18 at 18:34
  • @HansLundmark I thought $\frac{d^2u}{dx^2}=1$ could be treated as a fraction. Besides, $d^2u$ is the operator $d$ acting twice on $u$ and $dx^2$ is just $dxdx$, I believe, so to me the equation $d^2u=dxdx$ made sense – elcocodrilotito Nov 24 '18 at 18:38
  • But what is the operator $d$? And what does $dxdx$ mean? What kind of mathematical objects do these symbols denote? Until you can give precise answers to these questions, the expression is meaningless. – Hans Lundmark Nov 24 '18 at 18:55