# Solving the diff. eq. $\frac{dq}{dt}+ \left ( \frac{1}{3 \rho} \frac{d\rho}{dt} \right ) q =A(t)$

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.

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I noticed a misspelling in the title, and then another, but the system doesn't want there to be another question with the title "Solving a differential equation". I don't think that putting the equation in the title is such a bad idea, but that's a more serious change and I'll leave that up to you. –  Dylan Moreland Mar 29 '12 at 1:33

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

$$q(t)=\rho(t)^{-1/3}\int_0^t A(u)\rho(u)^{1/3}du.$$

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The problem is with right hand side of the equation as it is $A(t)$. –  sknandi Mar 29 '12 at 1:45
@sknandi: As you know, the original question had a typo (+ as an = sign) that mislead me. I have now addressed the current question. –  anon Mar 29 '12 at 2:04