Clarifying ODE/PDE Integration and Conditions

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SUMMARY

The discussion revolves around the integration of the equation $$ u \frac{ \partial u}{ \partial x} = \rho \frac{ d P}{ d x}$$ and its implications. The participants clarify that the correct formulation is $$ \rho P - \frac{1}{2} u^2 = const.$$ rather than the proposed positive formulation. It is established that the partial derivative can be treated as an ordinary derivative under specific conditions, and the integration process requires careful consideration of the signs involved in the equations.

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hey pf!

here's the question: $$ u \frac{ \partial u}{ \partial x} = \rho \frac{ d P}{ d x}$$ may i generally state $$ \rho P+1/2 u^2 = const. $$

the book does, and it seems the [itex]dx[/itex] cancels the [itex]\partial x[/itex] on both sides and we simply integrate through. this seems to be mathematically untrue. can someone confirm/reject this? also, what conditions would be necessary to have the above true (if it is indeed untrue generally)?
 
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I doubt very much that you book says exactly that! I suspect it says instead that
[itex]\rho P- (1/2)u^2= const[/itex]. (Notice the negative.)

The partial derivative is simply the ordinary derivative while treating other variables as if they were constants. What ever the other variable(s) in u might be, since they do not appear in the equation, this would be solved exactly as if it were
[tex]u\dfrac{du}{dx}= \rho\dfrac{dP}{dx}[/tex].

Now, you can treat [itex]du/dx[/itex] and [itex]dP/dx[/itex] as if they were ratios of differentials as we do in Ordinary Caculus.
 
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ahh yes, my mistake. the negative is definitely there. sorry. but thanks for answering the crux of the question
 

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