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?