Example comparison doubt

[VGNDISCPLN]

Hi everyone i hope you´re doing well. So i was studing by mi own transport phenomena and by following an example i found, that the equation:
becomes so i tried to solve de differential equation by making use of the method of integration by parts , and applying the boundary condition at which for x=0 vz(x)=0, so my result was the following:
[MATH] v_z (x)=-{\rho g cos \beta \over \mu_0} [{\delta x e^{\alpha x \over \delta}\over \alpha}-{\delta ^2 e^{\alpha x \over \delta} \over \delta ^2}+{\delta ^2 \over \alpha ^2}] [/MATH]Is it right or is it wrong or may be the same, but expressed in different way, can someone please answer this question?

 

[yoscar04]

Hi. I got a similar expression to yours, but in the second term inside the bracket the denominator is alpha^2.
I just used that the integral of xe^(s x)=e^s x (s x -1)/s^2

 

[VGNDISCPLN]

Hi. I got a similar expression to yours, but in the second term inside the bracket the denominator is alpha^2.
I just used that the integral of xe^(s x)=e^s x (s x -1)/s^2

yes it actually it is alpha^2 i made an error typing the equation

 

[JeffM]

Make life easy for yourself. I am assuming all those Greek letters are constants. Simplify by substitution of variables. The probability of error drops dramatically.

[MATH]\text {Let } A = \dfrac{\alpha}{\delta} \text { and } B = \rho g \delta^2 cos( \beta).[/MATH]
[MATH]\therefore \mu e^{-Ax} \dfrac{dv}{dx} = Bx \implies \dfrac{\mu}{e^{Ax}} * \dfrac{dv}{dx} = Bx.[/MATH]
That is obviously separable.

[MATH]\therefore \int dv = \dfrac{B}{\mu} * \int e^{Ax} x \ dx = \dfrac{B}{\mu} \left ( \dfrac{xe^{Ax}}{A} - \int \dfrac{e^{Ax}}{A} \ dx \right ) \implies\\ v = \dfrac{B}{\mu} * \left ( \dfrac{xe^{Ax}}{A} - \dfrac{e^{Ax}}{A^2} \right ) + C = \dfrac{B}{A \mu} * e^{Ax} \left (x - \dfrac{1}{A} \right ) + C.[/MATH]It is easy to check that.

[MATH]v' = \dfrac{B}{A\mu} * \left \{ e^{Ax} * 1 + Ae^{Ax} * \left ( x - \dfrac{1}{A} \right) \right \} = \\ \dfrac{B}{A\mu} (\cancel {e^{Ax}} + Ae^{Ax}x - \cancel {e^{Ax}}) = Bx * \dfrac{e^{Ax}}{\mu} \implies \\ \mu e^{-Ax} v' = Bx \ \checkmark[/MATH][MATH]0 = v(0) = \dfrac{B}{A \mu} * e^{A * 0} \left (0 - \dfrac{1}{A} \right ) + C \implies \\ 0 = \dfrac{B}{A \mu} * 1 \left (- \dfrac{1}{A} \right ) + C \implies C = \dfrac{B}{A^2 \mu} \implies \\ v = \dfrac{B}{A \mu} * e^{Ax} \left (x - \dfrac{1}{A} \right ) + \dfrac{B}{A^2 \mu} = \dfrac{B}{A \mu} * \left \{ e^{Ax} \left (x - \dfrac{1}{A} \right ) +\dfrac{1}{A} \right \}.[/MATH]

 

[VGNDISCPLN]

thank´s