Question

\[Q=Q_c+Q_p\]
\[ =e^{-10t}(c_1cos20t+c_2sin20t)+\frac{3}{125}\]
\[Q(0)=0\]
\[Q'(0)=0\]

Answer 1

how do I find c_1 and c_2?

Answer 2

@UnkleRhaukus

Answer 3

I'm almost done solving the problem, I just don't know how to use the initial charge and current (both are zero)

Answer 4

\[\large Q(0)=0 \quad \rightarrow \quad 0=e^{-10\cdot0}(c_1\cos20\cdot0+c_2\sin20\cdot0)+\frac{3}{125}\]
Hmm yah both are zero :D Interesting. You shouldn't any problem though, since cosine 0 will give you something useful.

As long as you understand where to plug in the 0's :D

Answer 5

cosine of zero is 1, and e^0 is one. 1=-3/125?

Answer 6

oooops

Answer 7

for c1? :D yah looks good!
c2 will be a bit more work.. since you need to find Q' :C

Answer 8

I meant c_1

Answer 9

Q'c=0

Answer 10

I mean Q'(0)=0

Answer 11

Yes so we need our equation in terms of Q' so we can plug in the 0's

Answer 12

\[\large Q(t)=e^{-10t}(c_1\cos20t+c_2\sin20t)+\frac{3}{125}\]\[\large Q'(t)=?\]

Answer 13

Bunch of product rule shinanigans :O

Answer 14

\[Q'=-10e^{-10t}(c_1cos20t+c_2sin20t)+e^{-10t}(-20c_1sin20t+20c_2cos20t)\]
\[\]

Answer 15

So now with the derivative, you might notice that you'll actually have 2 terms to worry about when you plug in zeros at this point - 2 terms that give you non-zero since the product rules produced a couple cosine terms.

So just a little bit more work to find c2! :)

Answer 16

\[Q'(0)=-10e^{-10(0)}(c_1)+e^{-10(0)}20c_2\]
\[Q'(0)=c_1+20c_2=0\]

Answer 17

do I have to take one more derivative

Answer 18

no c:

Answer 19

If there was a third unknown constant, and a third initial condition, then yah you might have to do that.

But that would only show up in a higher order DE :D

Answer 20

\[c_2=\frac{-c_1}{20}\]

Answer 21

And we already know what c1 is right? :D plug 'er in!

Answer 22

\[c_2=\frac{3}{2500}\]

Answer 23

From here,\[Q'(0)=-10e^{-10(0)}(c_1)+e^{-10(0)}20c_2\]Shouldn't this give us,\[\large c_2=\frac{c_1}{2}\]

Answer 24

oh I see I neglected the -10. makes sense =D
THanks sooo much!!!

Answer 25

heh :D