Thanks for that, as well as all the other pointers. I’ll do what you sand look more closely at how your code corresponds to the displayed maths.This site supports LaTeX. You can invoke delimiters for LaTex code by pressing the icon that looks a bit like a calculator
I know very little as well, this is a maths question from a maths textbook so understanding what it means is (thankfully) just extra flavor. I only put the RLC in the name in case there were electrical engineers here who would already be very familiar with it.I know no physics and so cannot help with that, but I’d use substitution of variables to simplify the math. I guarantee I’d screw up with the coefficients or exponents somewhere in that mess
Thank you so much for this! I would have never thought of putting it all in one fraction. That is definitely clearer than my response.I haven't slogged through it all, but I have a couple of comments, some of which you may already realize. You are given that [MATH]CR^2 > 4L[/MATH]. The [MATH]\lambda[/MATH] in your characteristic equation came out [MATH]\frac{-R \pm \sqrt{\frac{CR^2-4L} C}} {2L}[/MATH]. Note that if you choose the negative sign, the value of [MATH]\lambda[/MATH] is negative, and if you choose the positive sign, the value is still negative because the expression under the radical is less than [MATH]R^2[/MATH]. Therefore the complementary solution is a sum of two linearly independent negative exponentials and will fade to [MATH]0[/MATH] as [MATH]t\to\infty[/MATH]. So you just need the particular solution. Note that the right side is [MATH]\frac{dE}{dt} = V_0n\cos(nt)[/MATH]. You should be looking for a particular solution in the form [MATH]An\cos(nt) + Bn\sin(nt)[/MATH].
That’s where it’s all gone wrong for me and my 5 attempts, my particular solution is given at the end of my working and (as appears quite evidently) it won’t condense into the answer when I square them both, sum them and square root them. I am definitely looking for a slogger with inhuman organisation skills as I’m making some eroor(s) over and over again and getting confused by all these constantsSo you just need the particular solution
That is such a great idea. I just slogged through it with the of a teacher and it would have been so much easier if we'd done that. It's such a simple yet brilliant idea.Now set [MATH]v= -L\omega^2 + \frac 1 C[/MATH] and [MATH]u= R\omega[/MATH], so your two equations become:
[MATH]va - ub = 0\\ ua +vb =V_0\omega[/MATH]Solve these for [MATH]u,~v[/MATH] by determinants and simplify them. At the very end, plug in what [MATH]u[/MATH] and [MATH]v[/MATH] are and you should have it.