Undergrad What happens if you divide by a differential?

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SUMMARY

The discussion focuses on the implications of dividing a first-order separable differential equation (DE) by a differential, specifically examining the equation for pressure, P = ρgh, and its derivative forms. When dividing the differential dP by dt, the resulting equation, dP/dt = -ρg(dy/dt), raises questions about whether this leads to a partial differential equation (PDE) or remains solvable as a DE. The conclusion is that if y is expressed as a function of t, it allows for the determination of P(t), maintaining the relationship between P and y.

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  • Understanding of first-order separable differential equations
  • Familiarity with derivatives and differentials in calculus
  • Knowledge of partial differential equations (PDEs)
  • Basic principles of fluid mechanics, particularly pressure equations
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I've always wondered what happens when you divide a first order seperable DE by a differential? Does it then become a PDE? Is it still possible to solve it? If so how would one solve such a DE?

For example this:

Starting with this equation for pressure:
##P=ρgh## ; ## h = -y##
##P= -ρgy##
taking the derivative of this as ##Δy→0## we get

##\frac {dP} {dy} = -ρg##

and to solve this we'd normally go

##dP = -ρg dy##

What I'm wondering is what would happen if we were to now divide each side by a differential like let's say dt to get:

## \frac {dP} {dt} = -ρg \frac {dy} {dt} ##how would we end up solving this since we now essentially have two different "derivatives" on each side? would we just simply continue integrating each side through normally to obtain the functions P(t) and y(t) or would this instead be considered a PDE? in the form

## \frac {\partial P} {\partial t} = -ρg \frac {\partial y} {\partial t} ##
 
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If y is given as a function of t, this helps you to find P(t). But that ends up in the same relationship as P(y).

Grossly, dividing by a differential means dividing by a difference and then taking the limit ##\Delta \downarrow 0##.
 

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