# Challenging Second order Differential equation question (Model of RLC circuit)

#### Levido

##### New member
Hi. I’ve done part of this question but my particular integral looks far too complicated to simplify how the question (or the answer in the back I’ve seen) when expressed in harmonic form

question

My working:

If you’ve made it this far through my messy working than you so much already.

#### Levido

##### New member
Ill add the only helpful answer the book gives in the back, being that:

For absolute clarity, the 11 is the question number not the coefficient of K

On another topic, can anyone direct me to the post with how to code maths on this website?

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#### JeffM

##### Elite Member
Let me address the last question.

This site supports LaTeX. You can invoke delimiters for LaTex code by pressing the icon that looks a bit like a calculator.

People here will answer questions about LaTeX coding, but this site does not have a tutorial on it. One way to get an introduction to it is to hit “reply“ to a post with LaTeX code. The actual code will be displayed for you

$$\displaystyle L * \dfrac{d^2I}{dt^2} + R * \dfrac{dI}{dt} + \dfrac{1}{C} * I = \dfrac{dE}{dt}.$$

$$\displaystyle \lambda^2 + RL^{-1} \lambda + (CL)^{-1} = 0 \implies \lambda = \dfrac{- \dfrac{R}{L} \pm \sqrt{\dfrac{R^2}{L^2} - \dfrac{4}{CL}}}{2} = \dfrac{- \dfrac{R}{L} \pm \sqrt{\dfrac{CR^2}{CL^2} - \dfrac{4L}{CL^2}}}{2} = \dfrac{-R \pm \sqrt{\dfrac{CR^2 - 4L}{C}}}{2L}.$$

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.

#### Levido

##### New member
This site supports LaTeX. You can invoke delimiters for LaTex code by pressing the icon that looks a bit like a calculator
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.

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
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 fear I’ve done exactly what you said and messed up as things became messy

If anyone gets through this neatly, not only do you have my respect and gratitude but you should consider a job as a professional organizer (or maths helper )

Despite all the amazing helper’s normal format of coding all the maths, if anyone does attempt this in the end feel free to just send the working as an image or something, no coding unless you want to. I’ll appreciate it just as much

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#### LCKurtz

##### Full Member
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 $$\displaystyle CR^2 > 4L$$. The $$\displaystyle \lambda$$ in your characteristic equation came out $$\displaystyle \frac{-R \pm \sqrt{\frac{CR^2-4L} C}} {2L}$$. Note that if you choose the negative sign, the value of $$\displaystyle \lambda$$ is negative, and if you choose the positive sign, the value is still negative because the expression under the radical is less than $$\displaystyle R^2$$. Therefore the complementary solution is a sum of two linearly independent negative exponentials and will fade to $$\displaystyle 0$$ as $$\displaystyle t\to\infty$$. So you just need the particular solution. Note that the right side is $$\displaystyle \frac{dE}{dt} = V_0n\cos(nt)$$. You should be looking for a particular solution in the form $$\displaystyle An\cos(nt) + Bn\sin(nt)$$.
Also, to get the phase angle form use this:
$$\displaystyle a\sin \theta + b\cos\theta =\sqrt{a^2+b^2}\left( \frac a {\sqrt{a^2+b^2}}\sin\theta + \frac b {\sqrt{a^2+b^2}}\cos\theta \right ) = \sqrt{a^2+b^2}\left( \cos\alpha \sin\theta + \sin\alpha \cos\theta \right)$$
You don't need to figure out $$\displaystyle \alpha$$. You just need to note that $$\displaystyle K=\sqrt{a^2+b^2}$$. Hopefully it will all work out. Good luck with it.

#### Levido

##### New member
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 $$\displaystyle CR^2 > 4L$$. The $$\displaystyle \lambda$$ in your characteristic equation came out $$\displaystyle \frac{-R \pm \sqrt{\frac{CR^2-4L} C}} {2L}$$. Note that if you choose the negative sign, the value of $$\displaystyle \lambda$$ is negative, and if you choose the positive sign, the value is still negative because the expression under the radical is less than $$\displaystyle R^2$$. Therefore the complementary solution is a sum of two linearly independent negative exponentials and will fade to $$\displaystyle 0$$ as $$\displaystyle t\to\infty$$. So you just need the particular solution. Note that the right side is $$\displaystyle \frac{dE}{dt} = V_0n\cos(nt)$$. You should be looking for a particular solution in the form $$\displaystyle An\cos(nt) + Bn\sin(nt)$$.
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.

So you just need the particular solution
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 constants

Also, your full writing out of the putting the sum of sin and cos with different coefficient into harmonic form is something to behold. So much better than my textbook

Thank you for your help and well wishes, it means a lot LCKurtz

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#### LCKurtz

##### Full Member
Since this problem has caught my interest and I have a bit of time to kill, I'll give you an outline to follow below. I'm going to use $$\displaystyle \omega$$ instead of $$\displaystyle n$$ in the formulas.
We are looking for a particular solution of $$\displaystyle LI'' +RI' + \frac 1 C I = V_0\omega \cos(\omega t)$$ having the form
$$\displaystyle I = a\sin(\omega t) + b\cos(\omega t)$$. Its derivatives are:
$$\displaystyle I' = a\omega \cos(\omega t) - b\omega \sin(\omega t)$$
$$\displaystyle I'' = -a\omega^2\sin(\omega t) - b\omega^2\cos(\omega t)$$.

Put these in the equation:
$$\displaystyle L(-a\omega^2\sin(\omega t) - b\omega^2\cos(\omega t)) + R(a\omega \cos(\omega t) - b\omega \sin(\omega t)) +\frac 1 C (a\sin(\omega t) + b\cos(\omega t)) = V_0\omega\cos(\omega t)$$
Collect terms on the sines and cosines and set the coefficients on both sides equal. You should be able to get these two equations for $$\displaystyle a$$ and $$\displaystyle b$$.
$$\displaystyle (-\omega^2L +\frac 1 C) a -R\omega b = 0\\ R\omega a +( -\omega^2 L + \frac 1 C)b = V_0\omega$$
Now set $$\displaystyle v= -L\omega^2 + \frac 1 C$$ and $$\displaystyle u= R\omega$$, so your two equations become:
$$\displaystyle va - ub = 0\\ ua +vb =V_0\omega$$
Solve these for $$\displaystyle u,~v$$ by determinants and simplify them. At the very end, plug in what $$\displaystyle u$$ and $$\displaystyle v$$ are and you should have it.

#### LCKurtz

##### Full Member
Didn't notice until too late to edit, the last line should say solve for $$\displaystyle a$$ and $$\displaystyle b$$ then calculate $$\displaystyle K=\sqrt{a^2+b^2}$$.

#### Levido

##### New member
I like the changing of n to omega, maybe this is some sort of residual from SHM notation?

Now set $$\displaystyle v= -L\omega^2 + \frac 1 C$$ and $$\displaystyle u= R\omega$$, so your two equations become:
$$\displaystyle va - ub = 0\\ ua +vb =V_0\omega$$
Solve these for $$\displaystyle u,~v$$ by determinants and simplify them. At the very end, plug in what $$\displaystyle u$$ and $$\displaystyle v$$ are and you should have it.
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.

If anyone wants me to I'll post the full working. Otherwise, mods feel free to lock the thread if that's what this website does

#### nasi112

##### Full Member
That was a beautiful question. I regret that I came here late