I have a differential equation :
$ \frac{dq}{dt}+ \left ( \frac{1}{3 \rho} \frac{d\rho}{dt} \right ) q =A(t)$ with initial condition $q(0)=0$. I want solve the equation for $q$. Would you please help me in this regard.
I have a differential equation :
$ \frac{dq}{dt}+ \left ( \frac{1}{3 \rho} \frac{d\rho}{dt} \right ) q =A(t)$ with initial condition $q(0)=0$. I want solve the equation for $q$. Would you please help me in this regard.
Differential equations of the form $\dot{q}+f(t)q=g(t)$ allow for a sort of "reverse product rule" after multiplication by what's called an "integrating factor." We want the LHS to be (\mu q)'=\mu \dot{q}+\dot{\mu} q, so that we can integrate both sides directly and subsequently isolate $q$ from the other expressions.
For this to work, we must multiply by $\mu$ (this is what's in front of the $\dot{q}$ after multiplying after all), and we must also have $\dot{\mu}=f(t)\mu$ in order for the coefficient of $q$ to be consistent. This is a simple DE with solution $\mu = \mu(0) \exp \int_0^t f(\tau)d\tau$. Thus, after solving for the implicit constant via $q(0)=0$,
(\mu q)'=\mu g \quad\implies\quad q(t)=\frac{1}{\mu(t)}\int_0^t g(u)\mu(u)du.
Here $f(t)$ is a logarithmic derivative, $\frac{1}{3}\frac{d}{dt}\log\rho$, so we have $\mu=\sqrt[3]{\rho}$. Thus the final answer is
$q(t)=\rho(t)^{-1/3}\int_0^t A(u)\rho(u)^{1/3}du. $