I have an ODE equation to solve the velocity of the droplet, the equation reads:
$$ \rho_w u_d \frac{du_d}{dz} =( \rho_w- \rho_a)g-0.75c_d\rho_a\frac{(u_d-u_a)^2}{d_d} $$
I tried solving the equation considering it as an Initial value problem, I found no success as the equation reaches infinity.
$\rho_w$ =1000 $c_d$ =0.44 g= 9.81 $\rho_a$ = 1.25 $u_a$ = 4.55 intial condtion $u_d$ = 0 at z = 0 Need to find the variation along the $u_d$ along the height from Z= 0 till Z=4
$$ \rho_w u_d \frac{du_d}{dz} =( \rho_w- \rho_a)g-0.75c_d\rho_a\frac{(u_d-u_a)^2}{d_d} $$A key point in solving an equation is to write it on a standard form where is is easy to distinguish the unknown from the parameters.
In order to make clear this fishy equation, change the notations :
$u_d=y(z)\quad$ unknown function that you are looking for.
$\frac{( \rho_w- \rho_a)g}{\rho_w}=a\quad$ constant.
$\frac{0.75c_d\rho_a}{d_d \rho_w}=b\quad$ constant.
$u_a=c\quad$ constant. $$y\frac{dy}{dz}=a-b(y-c)^2$$ If you have done so, you would have recognize easily that the equation is a separable ODE which the way to solve is well known.
$$z=\int \frac{y\:dy}{a-b(y-c)^2}$$ Integrating gives the function $z(y)$ on closed form.
Then, one observe that inverting for $y(z)$ is not possible with standard functions. Thus you need numerical calculus to continue. Alternatively, solve directly the original ODE thanks to numerical calculus, which is the common way in practice.