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.