Question

solve Ld^2q/dt^2 + Rdq/dt + q/C = 0

Answer 1

are L and R and C variables? are we suppose to find a different equation or find a value for unknown?

Answer 2

L R C are constants @FaiqRaees

Answer 3

It is a linear second order ordinary differential equation with constant coefficients...

I need to find the general solution..

Answer 4

don't you use the characteristic equation
L r^2 +R r + 1/C =0
to find the two roots r1 and r2 (a bit ugly)
and the answer is a linear combination
q(t)= c1*exp( r1 t) + c2*exp(r2 t)

Answer 5

The solution is a combination of sine and cosine.

Answer 6

But my textbook says the solution is a decaying exponential multiplied by cosine @thomas5267

Answer 7

Oh crap. Guess \(e^{kx}\) as a solution. Plug it into your equation and solve for k. If the auxillary polynomial has two imaginary roots, then the solution would be the one suggested by your textbook if you rearrange the exponential functions to trig functions.

Answer 8

L r^2 +R r + 1/C =0

solving, i get
\[r = \dfrac{-R\pm \sqrt{R^2 - 4L/C}}{2L}\]

Answer 9

The solution is probably equivalent to the usual solution but it looks weird. In the underdamped case the textbook solution is the usual one but in other cases I don't think so.

Answer 10

you can factor out -R/2L

if the square root is negative (we need to assume values for the constants)
you get an exponential with imaginary component

Answer 11

Note that \(\omega_0^2=\dfrac{1}{LC}\).

Answer 12

\[
\begin{align*}
L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0\\
\frac{d^2q}{dt^2} + \frac{R}{L}\frac{dq}{dt} + \frac{q}{LC} &= 0\\
\frac{d^2q}{dt^2} + \frac{R}{L}\frac{dq}{dt} + \omega_0^2q &= 0\\
\end{align*}
\]
\[
k^2 + \frac{R}{L}k + \omega_0^2 = 0\\
\begin{align*}
k &= \frac{-\frac{R}{L}\pm \sqrt{\frac{r^2}{L^2}-4\omega_0^2}}{2}\\
k &= \frac{-\frac{R}{L}\pm 2\sqrt{\left(\frac{r}{2L}\right)^2-\omega_0^2}}{2}\\
k &= \frac{-\frac{R}{L}\pm 2i\sqrt{\left(\omega_0^2-\frac{r}{2L}\right)^2}}{2}\\
k &= -\frac{R}{2L}\pm i\sqrt{\left(\omega_0^2-\frac{r}{2L}\right)^2}
\end{align*}\\
q=c_1e^{-\frac{R}{2L} + i\sqrt{\left(\omega_0^2-\frac{r}{2L}\right)^2}}+c_2e^{-\frac{R}{2L} - i\sqrt{\left(\omega_0^2-\frac{r}{2L}\right)^2}}
\]
Rearrange to trig form and apply trig identities.

Answer 13

Awesome, thanks!

Answer 14

I just typed it up, so I may as well post it.
the two roots of the characteristic equation are
\[r= -\frac{R}{2L} \pm \frac{1}{2L} \sqrt{R^2 -\frac{4L}{C} } \]

putting the 1/2L inside the square root, that part becomes
\[ \sqrt{\left( \frac{R}{2L}\right)^2 - \frac{1}{LC} }\]
replace 1/LC with omega squared
factor out a -1 from the square root
\[ i \sqrt{ \omega^2-\left(\frac{R}{2L}\right)^2} \]
rename the "square root" stuff omega prime. the linear combination of solutions is
\[ q = e^{-\frac{R}{2L}t} \left(c_1 e^{i \omega ' t}+ c_2 e^{-i \omega ' t} \right) \]

the linear combination of exponentials can be written as a single phase shifted cosine (or sine)
(it takes a bit to write out how to do that)
\[ q = c e^{-\frac{R}{2L}t} \cos(\omega ' t + \phi) \]

Answer 15

I think we're assuming underdamped because \(\omega' \lt \omega\) can be inferred from the textbook... Makes so much sense, thanks a lot!