Recent content by Muskyboi

This is exactly what I was looking for. Thank you.

Is there a tool that can convert something like this: which when copied as plain text looks like this: \frac{4}{\pi}\sin\left(x\right)+\frac{4}{3\pi}\sin\left(3x\right)+\frac{4}{5\pi}\sin\left(5x\right)+\frac{4}{7\pi}\sin\left(7x\right)+\frac{4}{9\pi}\sin\left(9x\right) To this...

Yes, dividing the equation for mass of salt in the tank S(t) by the volume of water in the tank V(t) would give the concentration of salt in the tank. Doing so shows that the concentration approaches 1/5 as time increases.

v(t)=600+(9-6)t =600+3t 1500=600+3t therefore t=300 hrs when tank is full Cin=1/5(1 + cost) ds/dt=Rate in - rate out = CinRin - Cout*S(t)/V(t) =1/5(1 + cost)*9 - 6*S(t)/(600+3t) S(0)=5 ib Solving the first order linear ODE we get: https://www.desmos.com/calculator/l7iixzgyll therefore...

And what about the function for the current through the inductor iL(t)?

A2 is initial current or voltage to get A1 you must differentiate the equation with respect to time Find dv/dt or di/dt via KCL or KVL the dv/dt or di/dt will come from the equation for the capacitor current or inductor voltage (basically ohms law for inductors and capacitors ) now you sub...

OK, but How do I find A1 and A2 based upon the initial conditions?

Can't I just use v(t)=(A1+A2t)e^-αt since the circuit is critically damped (α=w0) and solve for A1 and A2 based upon initial conditions?

The inductor current curve and capacitor voltage curve should look like this after pulling the switch from which the circuit archived a steady state of 1 amp: link to circuit: http://tinyurl.com/y4hq6c6u