differential equation

[gonzales]

Hello everyone,

I solved a problem in chemical engineering and I reached to this equation. would you please help me to solve the equation. I attached its picture. here Vz is independent of Z-axis.

Thanks

 

[gonzales]

The correct equation is:

 

[topsquark]

It separates:
Let [math]C_A(x, z) = X(x) Z(z)[/math].

Then
[math]-D_{AB} \dfrac{ \partial ^2 C_A }{ \partial x ^2 } + V \dfrac{ \partial C_A }{ \partial z } - K C_A = 0[/math]
becomes
[math]-D_{AB} X'' Z + V X Z' - K X Z = 0[/math]
Divide both sides by X(x) Z(z) and let
[math]V \dfrac{Z'}{Z} = \lambda[/math] where [math]\lambda[/math] is a constant (wrt to x and z anyway)

Then
[math]D_{AB} \dfrac{X''}{X} + K = \lambda[/math]
-Dan

 

[gonzales]

It separates:
Let [math]C_A(x, z) = X(x) Z(z)[/math].

Then
[math]-D_{AB} \dfrac{ \partial ^2 C_A }{ \partial x ^2 } + V \dfrac{ \partial C_A }{ \partial z } - K C_A = 0[/math]
becomes
[math]-D_{AB} X'' Z + V X Z' - K X Z = 0[/math]
Divide both sides by X(x) Z(z) and let
[math]V \dfrac{Z'}{Z} = \lambda[/math] where [math]\lambda[/math] is a constant (wrt to x and z anyway)

Then
[math]D_{AB} \dfrac{X''}{X} + K = \lambda[/math]
-Dan

Thank you so much