# Homework Help: Differential equations assignment T3

1. Dec 2, 2013

### mathi85

Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.

A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the plates, V, is given by the differential equation:

CR(dV/dt)+V=E

a) Solve the equation for E given that when time t=0, V=0.
b) Evaluate voltage V when E=50V , C=10μF, R=200kΩ and t=1.2s

Solution:

a)
I want: (dV/dt)PV=Q

CR(dV/dt)+V=E /:(CR)
(dV/dt)+(1/CR)V=E/(CR)

P=1/(CR) and Q=E/(CR)

Integrating factor =e∫ P dt
∫ P dt=∫ 1/(CR) dt = ln(CR)t

IF=eln(CR)t=CRt

y(IF)=∫ (IF)*Q dx
∴V(IF)=∫ (IF)*Q dt
VCRt=∫ CRt*(E/(CR)) dt
VCRt=∫ tE dt
VCRt=(Et2)/2+c

General Solution:

VCRt=(1/2)Et2+c

t=0 when V=0

∴0=0+c
c=0

Particular Solution:

VCRt=(1/2)Et2 /*2
2VCRt=Et2 /:t2
E=(2VCRt)/(t2)

b)

VCRt=(1/2)Et2 /:CRt
∴V=(1/2)[(Et2)/(CRt)]

V=(1/2)[(50*1.22)/(10*10-6*200*103*1.2)]
V=30volts

2. Dec 2, 2013

### vela

Staff Emeritus
Where did this come from?

You didn't integrate correctly. R and C are constants.

This is wrong too. It's true that eln a=a, but that's not what you have here because of t.

3. Dec 2, 2013

### mathi85

Sorry! It should be:
"I want: (dV/dt)+PV=Q"

This is just to remind me what form the equation should have.

4. Dec 2, 2013

### mathi85

I'll try to get my head round this integration...

5. Dec 2, 2013

### mathi85

I really struggle with this integration involving 1/CR.
Is it just 1/(CR)*t and my IF will be e1/(CR)*t?

6. Dec 2, 2013

...et/(CR)

7. Dec 2, 2013

### vela

Staff Emeritus
Yes, that's right.

8. Dec 2, 2013

### mathi85

I understand that because CR are constants, not functions I cannot have it as ln.

9. Dec 2, 2013

### mathi85

Thnx! I'll do it again and post it asap.

10. Dec 2, 2013

### mathi85

Now it makes even less sense to me:

∫ P dt = ∫ 1/(CR) dt = t/(CR)
∴IF=et/(CR)

Vet/(CR)=∫ et/(CR)*E/(CR) dt

I'm not sure about the integration. If I treat E/CR as number then I can go:

Vet/(CR)= (CR)/E∫ et/(CR) dt

11. Dec 3, 2013

### vela

Staff Emeritus
This is fine:
$$Ve^{t/RC} = \int \frac{E}{RC}e^{t/RC}\,dt.$$ You're correct that E/RC is a constant so that you can pull it out of the integral; however, this isn't correct:
$$Ve^{t/RC} = \frac{RC}{E}\int e^{t/RC}\,dt.$$ Why did you flip the fraction over when you pulled it out of the integral? It should be
$$Ve^{t/RC} = \frac{E}{RC}\int e^{t/RC}\,dt.$$ Now use the substitution u = t/RC.

12. Dec 3, 2013

### mathi85

Sorry. I was tired last night and started making stupid mistakes.

I understand that now I have to use integration by parts method however in examples I've done so far I always had two functions of x, i.e. ∫ x*e2x dx
In this case I know that:

u=x
du/dx=1
dv/dx=e2x ∴ v=(1/2)e2x

and:
∫ u(dv/dx) dx = u*v-∫ v*(du/dx) dx

13. Dec 3, 2013

### mathi85

So if u=t/(CR), what is my dv/dt and v?

14. Dec 3, 2013

### vela

Staff Emeritus
Why are you using integration by parts?

15. Dec 3, 2013

### mathi85

Did you mean I should use a simple substitution?

Veu=E/(CR)∫ eu dt

16. Dec 4, 2013

### mathi85

I keep trying... This is what I have now:
Vet/(CR)=E/(CR)∫ e[1/(CR)]t dt
Vet/(CR)=E/(CR)*CRet/(CR)+c

17. Dec 4, 2013

### vela

Staff Emeritus
Looks good. Keep going.

18. Dec 12, 2013

### mathi85

Right I come to the point where:

Ve(1/(CR))t=Ee(1/(CR))t+c /:e(1/(CR))t
V=E+c/(e(1/(CR))t)

if I apply t=0 when V=0

0=E+c
c=-E

and Part. Sol.

V=E+[-E/e(1/(CR))t]

Where am I going wrong?

19. Dec 12, 2013

### mathi85

I'm getting that E=V+E/e(1/(CR))t

This cannot be right can it?

20. Dec 12, 2013

### vela

Staff Emeritus
That's correct. Why do you think anything's wrong?

21. Dec 12, 2013

### mathi85

It just didn't look right to me.
Thank you very much for your help!