Integrating a First Order Differential Equation (The West Equation) I am currently doing a project about Growth and have found this really interesting Math Model by Dr. Geoffrey West et al in 2001 while researching. The paper can be found at this link. 
I was wondering how does one integrate the differential equation (as in the paper)
$$ \frac{dm}{dt} = am^{0.75} \left [1- \left(\frac{m}{M}\right)^{0.25}\right]$$
where $m$ is a function in terms of $t$ and $a, M$ are constants. 
The end result (the integrated expression) presented by the paper is:
$$\left(\frac{m}{M}\right)^{0.25} = 1 - \left[1 - \left(\frac{m_0}{M}\right)^{0.25}\right] e^{-at/(4M^{0.75})}$$
where $m_0$ is another constant (in the context of the problem, the initial mass of organism when $t = 0$). 
I have tried solving by the integrating factors method but I couldn't express it in the form given... Anyone has any ideas?  
 A: The solution given by wdacda treats the equation as a Bernoulli equation. It can also be solved as an equation in separeted variables:
$$
\frac{dm}{m^{0.75}\Bigl[1- \Bigl(\frac{m}{M}\Bigr)^{0.25}\Bigr]}=a\,dt.
$$
Integrating between $0$ and $t$
$$
\int_{m_0}^{m}\frac{dz}{z^{0.75}\Bigl[1- \Bigl(\frac{z}{M}\Bigr)^{0.25}\Bigr]}=a\,t.
$$
The integral is easily computed by partial fractions.
A: The solution of the linear initial value problem 
$$
4 y'(t) = 1 - y(t), \quad y(0) = y_0,  
$$
is
$$
y(t) = 1+(y_0 - 1)e^{-t/4}. 
$$
Substituting $y(t) = x(t)^{1/4}$ we get that the solution of the initial value problem
$$
x'(t) = x(t)^{3/4} \bigl(1 - x(t)^{1/4}\bigr), \quad x(0) = x_0 \gt 0  
$$
is
$$
x(t) = \bigl(1+(x_0^{1/4} - 1)e^{-t/4}\bigr)^4. 
$$
Substituting $x(t) = \alpha\, w(\beta t)$, with $\alpha, \beta \gt 0$ we get that the solution of the initial value problem
$$
w'(\beta t) = \alpha^{-1/4} \beta^{-1} w(\beta t)^{3/4} \bigl(1 - \alpha^{1/4} w(\beta t)^{1/4}\bigr), \quad w(0) = w_0 \gt 0
$$
is
$$
w(\beta t) = \alpha^{-1} \bigl(1+(w_0^{1/4} \alpha^{1/4} - 1)e^{-t/4}\bigr)^4.
$$
Now replacing $\beta t$ by $t$ we get that the solution of the initial value problem
$$
w'(t) = \alpha^{-1/4} \beta^{-1} w(t)^{3/4} \bigl(1 - \alpha^{1/4} w(t)^{1/4}\bigr), \quad w(0) = w_0 \gt 0
$$
is
$$
w(t) = \alpha^{-1} \bigl(1+(w_0^{1/4} \alpha^{1/4} - 1)e^{-t/(4\beta)}\bigr)^4.
$$
Finally, setting 
$$
\alpha = M^{-1} \quad \text{and} \quad \beta = a^{-1} M^{1/4},  
$$
where $a, M \gt 0$ we get that the solution of the initial value problem
$$
w'(t) = a\, w(t)^{3/4} \bigl(1 - M^{-1/4} w(t)^{1/4}\bigr), \quad w(0) = w_0 \gt 0
$$
is
$$
w(t) = M \bigl( 1 + (w_0^{1/4} M^{-1/4} - 1)e^{-at/(4M^{1/4})}\bigr)^4.
$$
So, the solution that you give is not correct. 
