Help with integrating differential equation

[teeeeee]

Hi,

Could someone help me see how the solution of the equation \frac{1}{\mu} \frac{\partial p}{\partial z} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \frac{\partial v}{\partial \rho})

is v = \frac{1}{4\mu} \frac{\partial p}{\partial z} \rho^{2} + C_{1}ln(\rho) +C_{2}

Thank you

teeeeee

 

[teeeeee]

Hi,

Could someone help me see how the solution of the equation \frac{1}{\mu} \frac{\partial p}{\partial z} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \frac{\partial v}{\partial \rho}) is v = \frac{1}{4\mu} \frac{\partial p}{\partial z} \rho^{2} + C_{1}ln(\rho) +C_{2}

Thank you

teeeeee

 

[teeeeee]

Anyone? Please help

 

[payumooli]

assume partial p over partial z to be some constant 'k'
expand the right side you now have a second order linear differential equation with a coeff
something like
(D^2 + D)v = (k/mu)*rho
remember rho is not a constant so find the transient solution
the sum of the steady state and transient solution is your answer
hope this helped

ps: i am lazy to use symbols

 

[LCKurtz]

Hi,

Could someone help me see how the solution of the equation \frac{1}{\mu} \frac{\partial p}{\partial z} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \frac{\partial v}{\partial \rho})

is v = \frac{1}{4\mu} \frac{\partial p}{\partial z} \rho^{2} + C_{1}ln(\rho) +C_{2}

Thank you

teeeeee

It isn't clear what variables depend on what, but you can check this. I will use subscripts for partials. Let's begin by calling

w = \frac{1}{\mu} \frac{\partial p}{\partial z}.


So you have

\rho w = (\rho v_{\rho})_{\rho}. Integrate with respect to \rho:

\frac {\rho^2} 2 w + C =\rho v_{\rho},\ v_{\rho}=\frac {\rho} 2 w + \frac C {\rho}

Integrate again:

v =\frac {\rho^2} 2 w + C\ln \rho + D = \frac {\rho^2} 2\frac{1}{\mu} \frac{\partial p}{\partial z}+ C\ln \rho + D

You can check it. Something is off by a factor of 1/2.