- #1
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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} ##
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} ##