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Let's say we have the following DE:

$ \frac{dy}{dx} = x $

1. That's how It could be solved:

$dy = x dx$

$\int{dy}=\int{xdx}$

$y=\frac{1}{2}x^2+c$

Is it mathematically correct to separate $dy$ and $dx$ ? Or it would appear as the derivative of $y$ with respect to $x$ is a fraction which is not true.

2. Another method of writing the solution can be:

$\int\frac{dy}{dx} dx=\int{xdx}$

$\int{dy}=\int{xdx}$

$y=\frac{1}{2}x^2+c$

That's integrating the DE with respect to x. Is this mathematically correct? It seems like the $dx$ cancels each other in the LHS, but we are not allowed to treat this as a fraction.

My Questions are:

  • Which of these ways of writing the solution is the most accurate?
  • And which of these 2 is wrong in terms of following mathematical structure?
  • When we integrate both sides of the equation, can we just put the integral sign without integrating with respect to a variable? (for example in case 1)
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    $\begingroup$ Third method should be $\int y'\,dx.$ If you made that change, all three methods are equally valid. $\endgroup$ – Adrian Keister Oct 1 '19 at 17:12
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    $\begingroup$ It is probably more correct to write $\int {dy(x) \over dx} dx = \int x dx$... $\endgroup$ – copper.hat Oct 1 '19 at 17:16
  • $\begingroup$ To justify the first method in full rigor, one needs to develop the theory of differential forms. But the upshot is that, all the manipulations in your solution can be validated. $\endgroup$ – Sangchul Lee Oct 1 '19 at 17:25
  • $\begingroup$ The integral sign is very important as it defines the integration. It also tells us if it indefinite or definite. $\endgroup$ – Sam Oct 1 '19 at 18:00
  • $\begingroup$ Related: math.stackexchange.com/questions/27425/… $\endgroup$ – Hans Lundmark Oct 2 '19 at 4:03
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The first two are O.K. - the second being a more rigorous statement.
The third is wrong. $\int y'dy \ne y$, since $y'\ne 1$.

I don't quite get what the third question asks.

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It is separable equation where $ y= y(x)$. Consider the problem $$\frac{dy}{dx}=F(x)*Q(y)$$ where $F(x)$ depends only of $x$ and $Q(y)$ only of $y$. If $Q(y) \neq 0$ we can rewrite this as $$\frac{y'(x)}{Q(y(x))}= F(x)$$ $$\int_{x_0}^x{\frac{y'(t)}{Q(y(t))}dt}=\int_{x_0}^x{F(t)dt}$$ Then $y(t)=s$ We will get $$\int_{y(x_0)}^{y(x)}{\frac{ds}{Q(s)}} = \int_{x_0}^x{F(t)dx}$$

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When your diffferential equation is separable you end up with a differential form $$ f(y)dy = g(x) dx$$

Since you are assuming taht $$y=y(x)$$ you may integrate both sides with respect to $x$ but the LHS has gone through a change of variable so we get $$\int f(y(x))y'(x)dx = \int f(y)dy = \int g(x) dx$$

Thus the integration is valid.

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