# Does this equation makes sense? If so, how to solve it?

• Dec 2nd 2009, 10:06 AM
arbolis
Does this equation makes sense? If so, how to solve it?
I'm wondering if the following equation makes sense, and how to solve for $\displaystyle q(t)$.

$\displaystyle -\frac{dt^2}{LC}=\frac {d^2q}{q}$.

I found this equation when trying to solve the following problem : http://www.physicshelpforum.com/phys...pacitance.html.
I wanted to find $\displaystyle V(t)$. I realize the solution should be sinusoidal, so something like $\displaystyle V(t)=V_{max} \sin (\omega t + \phi)$.
Regarding the problem, when the capacitor halved its energy, I've found out that $\displaystyle V(0.0021s)=\frac{V_{max}}{\sqrt 2}$. So I've found the boundary values and if I find $\displaystyle V(t)$ I'm done.
If I find $\displaystyle q(t)$, I have that $\displaystyle V(t)=\frac{q(t)}{C}$ and so I'm done.
I used Kirchhoff's law of voltage to find $\displaystyle -\frac{dt^2}{LC}=\frac {d^2q}{q}$. I might have done errors though.

Thank you for any help!
• Dec 2nd 2009, 01:46 PM
CaptainBlack
Quote:

Originally Posted by arbolis
I'm wondering if the following equation makes sense, and how to solve for $\displaystyle q(t)$.

$\displaystyle -\frac{dt^2}{LC}=\frac {d^2q}{q}$.

I found this equation when trying to solve the following problem : http://www.physicshelpforum.com/phys...pacitance.html.
I wanted to find $\displaystyle V(t)$. I realize the solution should be sinusoidal, so something like $\displaystyle V(t)=V_{max} \sin (\omega t + \phi)$.
Regarding the problem, when the capacitor halved its energy, I've found out that $\displaystyle V(0.0021s)=\frac{V_{max}}{\sqrt 2}$. So I've found the boundary values and if I find $\displaystyle V(t)$ I'm done.
If I find $\displaystyle q(t)$, I have that $\displaystyle V(t)=\frac{q(t)}{C}$ and so I'm done.
I used Kirchhoff's law of voltage to find $\displaystyle -\frac{dt^2}{LC}=\frac {d^2q}{q}$. I might have done errors though.

Thank you for any help!

It looks like you have tried separating variables for what is essentially a second order homogeneous constant coefficient linear ODE. That is not the way to solve this, you normally use a trial solution of the form $\displaystyle q(t)=e^{\lambda t}$ which will give a quadratic for $\displaystyle \lambda$. Alternatively recognise the ODE as that of SHM and just write down the solution.

CB
• Dec 2nd 2009, 01:47 PM
arbolis
Quote:

Originally Posted by CaptainBlack
It looks like you have tried separating variables for what is essentially a second order homogeneous constant coefficient linear ODE. That is not the way to solve this, you normally use a trial solution of the form $\displaystyle q(t)=e^{\lambda t}$ which will give a quadratic for $\displaystyle \lambda$. Alternatively recognise the ODE as that of SHM and just write down the solution.

CB

Thank you very much!