Differential equation - help How should I tackle this differential equation
$\frac{d \ln{y(t)}}{d \ln{t}} = \alpha (1 - \frac{p(t)}{y(t)})$
in the unknown function $y(t)$ ?
Separation of variables maybe? Thanks to anyone who will help me!
 A: Since $d(\ln y)=dy/y$ and $d(\ln t)=dt/t$, the equation can be written as
$$
\frac{t}{y}\,\frac{dy}{dt}=\alpha\,\Bigl(1-\frac{p}{y}\Bigr).
$$
This is equivalent to
$$
\frac{dy}{dt}=\frac{\alpha}{t}\,y-\frac{\alpha\,p}{t},
$$
which is a linear equation.
A: Using the chain-rule, we can write 
$$\begin{align}
\frac{d\log y(t)}{d\log t}&=\frac{d\log y(t)}{dy(t)}\frac{d  y(t)}{dt}\,\frac{d t}{d\log t}\\\\
&=\frac{ty'(t)}{y(t)}
\end{align}$$
Then, the ordinary differential equation (ODE) can be written
$$\frac{ty'(t)}{y(t)}=\alpha\left(1-\frac{p(t)}{y(t)}\right)$$
which after multiplying both sides by $y(t)/t$ and rearranging terms reveals 
$$y'(t)-\frac{\alpha}{t}y(t)=-\alpha p(t)/t$$ 
To solve this ODE, we use the integrating factor $\mu(t) = t^{-\alpha}$.  Aside, the integrating factor can be found by noting that $(\mu y)'/\mu =y'+(\mu '/\mu)y$, setting $\mu ' /\mu =-\alpha t^{-1}$, and solving for $\mu$.  Proceeding, we find that 
$$y(t)=(t/t_0)^{\alpha}y(t_0)-\alpha t^{\alpha} \int_{t_0}^t t'^{-\alpha-1}p(t')dt'$$
