# Homework Help: Completely confused about differential equation notation

1. Apr 11, 2010

### darryw

1. The problem statement, all variables and given/known data
I just finished solving about 10 different DEs, but this one has me totally confused. It says:
(R)dQ/dt + (Q/C) = 0, with initial condition Q(0) = Q_0?

Most of the problems ive done thus far have looked like: y' + (2/t)y = (cost)/(t^2), with initial cond y(pi) = 0 and t>0 for example.

So then the RdQ/dt problem is confusing me. I mean, dQ, dt, R, C and Q.. How am I supposed to solve this? Please clarify what they are asking.. I can do these problems but there seems to be 3 variables in this problem?? please help, thanks

Also, shouldnt the correct notation put the R in front of dQ/dt? so it should look like dQ/dt (R) + Q(1/C) = 0 ??

2. Relevant equations
Specifically, if you could show me how this: (R)dQ/dt + (Q/C) = 0 , is similar to this: y' + p(x)y = q(x) .. I know that y' is the same thing as dy/dx, so is dQ/dt the same as saying Q' ? and if so, then R is the variable right? If that is true then what is C??

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Apr 11, 2010
2. Apr 11, 2010

### phyzguy

In this equation, R and C are constants, really no different than the constant '2' in your example: y' + (2/t)y = (cost)/(t^2). Treat them as constants and find Q as a function of t. Your answer for Q(t) will contain R and C.

3. Apr 11, 2010

### rock.freak667

I am inclined to think that R and C are constants as in a circuit, R is resistance and C is capaciatnce.

But it becomes the same as (R)Q'+(1/C)Q=0 with R and C as constants.

4. Apr 11, 2010

### darryw

Thanks alot.. this really helped.. i just get accustomed to a pattern and i couldnt see the equation any other way.

5. Apr 12, 2010

### darryw

Can someone please check my work?? ( i put my specific questions in parenthesis after each step) Thanks.

(R)dQ/dt + (Q/C) = 0, with initial condition Q(0) = Q_0

(R)Q' + (1/C)Q = 0

Q' + (1/CR)Q = 0 (is it always necessary to isolate Q' like this before i proceed with solving?)

mu(x) = e^(integ(1/CR))

mu(x) = CR (natural log cancels the exponent, right?)

(then i multiplied by my integrating factor...

(CRQ)' = integ (0)

then i integrate with respect to t...

CRQ = t + c

Q = (1/CR)(t+c)

apply initial conditions of Q(0) = Q_0

Q_0 = (1/CR)(0+c)

if this is all correct so far, then i am unclear as to how to finish.. i mean, do i solve for c and then apply that to the origical DE?

6. Apr 12, 2010

### rock.freak667

Well we try to make it into a recognizable form

Q'+P(t)Q= R(t) or using whatever symbols you are comfortable with

Remember C and R are constants, so μ=∫(1/CR) dt = (1/CR) ∫1 dt = ?

That is your integrating factor. Now you multiply by μ throughout in the DE.

Remember that in the circuit, resistance R and capacitance C is not changing with time or anything.

7. Apr 12, 2010

### darryw

OK so integrating factor is actually mu(x) = (1/CR) integ 1 dt = t/CR
so applying this new factor...
((t/CR)Q)' = (0)

integrate both sides..

(t/CR)Q = 0 (because integral of zero is zero with no c leftover)

but when i divide through by (t/CR), i end up with :

Q = 0

this doesnt seem right???

Last edited: Apr 12, 2010
8. Apr 12, 2010

### rock.freak667

remember that if you differentiate a constant you will get zero. So is the right side really just zero?

9. May 13, 2010

### MisterMan

Sorry to post so late, but I came across this in a google search and I think it will benefit others if I post. I believe I have the general solution for the problem, I cannot however plug in the initial condition as I'm not sure what Q_0 means in relation to the question, but anyway here is what I calculated :

$$R\frac{dQ}{dt} + \frac{1}{C}Q = 0$$

Dividing through by R gives :

$$\frac{dQ}{dt} + \frac{1}{RC}Q = 0$$

Now as posted by other people, $$\frac{1}{RC}$$ is a constant just like 2,3 and any integer you can think of. So when finding the integrating factor, it is necessary to remember that we are integrating with respect to t, not R or C :

$$Integrating Factor = e^{\int{\frac{1}{RC}}dt} = e^{{\frac{1}{RC}}t}$$

Now we multiply both sides of the differential equation by the integrating factor:

$$e^{{\frac{1}{RC}}t}\frac{dQ}{dt} + e^{{\frac{1}{RC}}t}\frac{1}{RC}Q = 0$$

Now we integrate both sides with the respect to t :

$$e^{{\frac{1}{RC}}t}Q = \int 0\,dt = c$$

where $$c$$ is the constant of integration.

Finally, we get the equation in terms of Q by dividing both sides by the integrating factor:

$$Q = \frac{c}{e^{{\frac{1}{RC}}t}}$$

Hope this helps, bye for now.