Differential equations assignment T3

In summary: I don't know what I would have done without you! In summary, the conversation was about a student seeking help with a task involving solving a differential equation and evaluating a voltage. The student struggled with integration and made some mistakes, but with the help of an expert, they were able to correctly solve the equation and find the voltage. The conversation ended with the student expressing gratitude for the expert's help.
  • #1
mathi85
41
0
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.
Thank you in advance for your time.

Task 3:

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
 
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  • #2
mathi85 said:
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.
Thank you in advance for your time.

Task 3:

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
Where did this come from?

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

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

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
 
  • #3
Sorry! It should be:
"I want: (dV/dt)+PV=Q"

This is just to remind me what form the equation should have.
 
  • #4
I'll try to get my head round this integration...
 
  • #5
I really struggle with this integration involving 1/CR.
Is it just 1/(CR)*t and my IF will be e1/(CR)*t?
 
  • #6
...et/(CR)
 
  • #7
Yes, that's right.
 
  • #8
I understand that because CR are constants, not functions I cannot have it as ln.
 
  • #9
Thnx! I'll do it again and post it asap.
 
  • #10
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
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
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
So if u=t/(CR), what is my dv/dt and v?
 
  • #14
Why are you using integration by parts?
 
  • #15
Did you mean I should use a simple substitution?

Veu=E/(CR)∫ eu dt
 
  • #16
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
Looks good. Keep going.
 
  • #18
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
I'm getting that E=V+E/e(1/(CR))t

This cannot be right can it?
 
  • #20
That's correct. Why do you think anything's wrong?
 
  • #21
It just didn't look right to me.
Thank you very much for your help!
 

FAQ: Differential equations assignment T3

1. What are differential equations?

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are often used in modeling natural phenomena and physical systems.

2. What is the purpose of the T3 assignment on differential equations?

The T3 assignment on differential equations is designed to test your understanding and application of the concepts and techniques learned in class, and to help you develop problem-solving skills in this area of mathematics.

3. What are some common applications of differential equations?

Differential equations have many applications in science, engineering, and economics. Some examples include modeling population growth, predicting the spread of diseases, and describing the behavior of electrical circuits.

4. What are the different methods for solving differential equations?

There are several methods for solving differential equations, including separation of variables, substitution, and using integrating factors. The specific method used depends on the type and complexity of the equation.

5. How can I improve my skills in solving differential equations?

Practice is key when it comes to improving your skills in solving differential equations. It is important to understand the fundamental concepts and techniques, and to work through a variety of problems to gain experience and build confidence.

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