Question

How is this true?

Answer 1

\[\frac QC\frac{dQ}{dt}+LI\frac{dI}{dt}=0\implies\frac QC+L\frac{d^2Q}{dt^2}=0\]

Answer 2

let I = dQ/dt

Answer 3

I think I saw the relationship:
\[\Large I=\frac{dQ}{dt} \]
once, if I remember correctly.

Answer 4

I understand that\[\frac{dI}{dt}=\frac{d^2Q}{dt^2}\]and that's just physics,but can you "factor out" the other \(\frac{dQ}{dt}\) like that??

Answer 5

don't you get a separate solution by letting dQ/dt=0 if you try something like that?
or is it just a matter that the solution to that must be a constant, and so gets absorbed into other constants?

Answer 6

I would say physical wise the solution where the intensity is zero just is not interesting enough, but mathematically you're right, it appears to be a trivial solution of the above. If it get's factored out - or especially divided by - however;
\[\Large I\neq0 \]
has to hold

Answer 7

This represents an LC circuit,and so to me the solution to dI/dt=0 should be important to the behavior of the system, yet the book I has simply says they are the same.

Answer 8

capacitor-inductor circuit...but I was more concerned with the math,which is why I posted it here.

Answer 9

I suppose if the solution to dQ/dt=0 is Q(t)=K where K is some constant, then we just have\[K\frac QC+KL\frac{dI}{dt}=0\] and K drops out... is that legit?

Answer 10

but yeah, if I=0 ...

Answer 11

then it's trivial, you are right
ok, had to think it through, thanks @Spacelimbus

Answer 12

You're welcome @TuringTest, I tried to look through my textbook, I once saw an example where they did just add additional solutions with a specific notation of a fundamental solution set. However, I believe in this case it's best to treat is as a trivial solution.