I'm currently doing a summer school course of applied mathematics for engineering and we've been given an assignment in First Order Ordinary Differential Equations. I have never really encountered them before, and my overall maths level is not really that good (hence the attempt at summer school), as an example to work with, we were given the following equation, where everything is a function of t except for R and C which are constants.
$\frac{dP}{dt}+\frac{P-P_{out}}{RC}=\frac{Q_{in}}{C}, P(t=0) = P_{0}$
We are told to show that the solution for pressure P(t) is the following equation:
$P=P_{out}+(P_{0}-P_{out})e^{-\frac{t}{RC}}+\frac{e^{(-\frac{t}{RC})}}{C}\int_{0}^{t}Q_{in}(\hat{t})e^{\frac{\hat{t}}{RC}}d\hat{t}, t\geq 0$
I'm pretty sure I'm supposed to go about solving this using integration factors but I'm finding it difficult to get the factor I'm supposed to multiply by.
I'm sure I'm missing something obvious here, but nevertheless, no matter how hard I try to find a similar equation to relate to, I haven't figured it out, any help would be greatly appreciated.
Best Answer
It is a first order differential equation in variables $t$ and $P$. First rewrite the equation as $\frac{dP}{dt}+\frac{1}{RC}P=\frac{Q_{in}(t)}{C}\frac{P_{out}}{C}$. The integrating factor is $e^{\frac{t}{RC}}$. Then multiply the equation by this factor it becomes $\frac{d}{dt}(e^{\frac{t}{RC}} P)=(\frac{Q_{in}(t)}{C}\frac{P_{out}}{C})e^{\frac{t}{RC}}$. Integrating both sides and using the initial condition $P(0)=P_0$ you get the solution.