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If I take a very simple differential equation $\frac{dy}{dx} = y$ and apply the separation rule to get to $\frac{dy}{y} = dx$, I obtain $ln|y|=x+c$, and thus $y = ce^x$.

However, in case I separate the terms differently to gain $dy=y dx$ and, as in the previous instance, integrate both sides of the equation, the result will be $y=yx+c$, which is clearly not the same and also it is quite distinctly wrong (at least considering the result WolframAlpha provided).

I have just began learning all this stuff about differential equations and this leaves me baffled - the latter method seems correct to me, and yet it yeields a wrong result. Could someone please clarify where exactly the mistake is made, i.e. which step is the "illegal" one in the second approach? Have I made any false assumptions?

Thanks in advance.

edit: Thanks for prompt answers, the problem was treating y as a constant while integrating it.

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    $\begingroup$ y is a function of x; that is y=y(x) $\endgroup$ – randomgirl Feb 7 '15 at 23:46
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I can't comment (don't have enough reputation), but I believe that the mistake is in the part where you take:

$$ dy=ydx $$

and you integrate it to get

$$ y=yx+c $$

The left hand side is fine, but when you integrate $y$ with respect to $x$, since $y$ is a variable function, you can't treat it like a 'constant'. Thus the integration to get $yx+c$ is invalid.

Another way to look at it I guess is:

take $y=yx$ and differentiate it. If $y$ and $x$ are both functions. This gives:

$$\frac{dy}{dx} =y'x+yx'$$

while if $y$ is a constant (like you treated it) it would give:

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

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  • $\begingroup$ Oh my I see that now, many thanks. Quite a stupid overlooking :x $\endgroup$ – Nihilus Feb 8 '15 at 0:02
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if you integrate $dy = ydx$ what you get is $y = \int y \, dx + C$ not as you have $y = xy + C$

the reason is $y$ is a function of $x$ and you are trying to determine that. the function $e^x$ will satisfy the $e^x = \int e^x dx + C$ as you can see.

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