Solve DE using integ factor, please check work thanks

In summary: So your constant can be written as e^c, which gives you the same form as the integrating factor solution. In summary, the conversation discusses solving a differential equation using both the integrating factor method and the separable method. The integrating factor method involves using an integrating factor to simplify the equation, while the separable method involves separating the variables and integrating. The two methods result in the same solution, but the integrating factor method may provide a more efficient solution. Initial conditions are also discussed, and it is determined that the constant in the general solution can be determined by plugging in the initial condition.
  • #1
darryw
127
0

Homework Statement



R dQ/dt + Q/C = 0 ...Q(0) = Q_0

dQ/dt = - Q/RC

integrating factor: mu(x) = e^integ(1/RC)
mu(x) = e^(t/RC)

(e^(t/RC) Q)' = integ 0

e^(t/RC) Q = t + c

Q = te^-(t/RC) + ce^-(t/RC)

then apply initial conditions of Q(0) = Q_0

Q_0 = c

the last part seems sort of weird to me?? thanks

Homework Equations





The Attempt at a Solution









 
Physics news on Phys.org
  • #2
By simply plugging your solution into the differential equation you would've noticed that your solution is wrong. Why are you using an integrating factor on a separable differential equation?

(e^(t/RC) Q)' = integ 0

e^(t/RC) Q = t + c

integ 0 != t+c
 
  • #3
yes i noticed it can be solved as separable, but shouldn't it give same answer using integrating factor?

my mistake was integral of zero is not t, it is just the arb constant, c
so...

(e^(t/RC) Q)' = integ 0

e^(t/RC) Q = c

Q = c^e^-(t/RC)

is this correct (so far) ? thanks
 
  • #4
You have raised the constant to the power of the exponent. I reckon that is a typo? If it is a typo then yes your answer is correct.
 
  • #5
thanks.. yup, typo..should be..
Q = ce^-(t/RC)

how exactly do i apply the init cond though?
it says Q(0) = Q_0, so doesn't that mean:

Q_0 = ce^-(0/RC)

Q_0 = c ?

this is what confuses me, because i am used to initial conditions being something like y(0) = 1, for example, then i just plug in the y value and x value to solve for the constant.. but when it says Q(0) = Q_0, i don't know what the problem is asking??
thanks.
 
  • #6
You should think of it like this. You have a solution [itex]Q(t)[/itex] and an initial condition [itex]Q(0)=Q_0[/itex]. This tells you that your general solution [itex] Q(t)[/itex] must be equal to [itex]Q_0[/itex] when [itex]t=0[/itex]. The only way for your general solution to meet that requirement is to set [itex]c=Q_0[/itex] as you already did.
 
  • #7
ok i think i got it.
so when init conditions, answer is :
Q = Q_0e^-(t/RC)

So if i solve as separable, i get this...R dQ/dt + Q/C = 0 ...Q(0) = Q_0

dQ/dt = -Q/C

(1/Q)dQ = -(1/C)dt

then integrate both sides...

ln|Q| = -ln|C|+c

exponentiate both sides...

Q = -C + e^c

this last part looks wrong though.. ??
thanks
 
  • #8
ok i think i got it.
so when init conditions, answer is :
Q = Q_0e^-(t/RC)

Correct.

R dQ/dt + Q/C = 0 ...Q(0) = Q_0

dQ/dt = -Q/C

(1/Q)dQ = -(1/C)dt

then integrate both sides...

ln|Q| = -ln|C|+c

You forgot the R in your first step.

Your second step is correct if you add the R.

In the 4th step you have integrated over C, but C is a constant. You need to integrate with respect to t.
 
  • #9
R dQ/dt + Q/C = 0 ...Q(0) = Q
dQ/dt = -Q/RC

(1/Q)dQ = -(1/RC)dt

now integrate..

ln|Q| = -(t/RC)+c

exponentiate...

Q = e^(-(t/RC)+c)

but this is slightly different from the result i got with integ factor??
 
  • #10
No it is the same expression, because [itex]e^c=constant[/itex].
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates the rates of change of one or more variables. It involves derivatives, which represent the rate of change of a function with respect to its variables.

2. What is an integrating factor?

An integrating factor is a function used to solve a differential equation by multiplying both sides of the equation. It helps to make the equation easier to solve by reducing the number of terms or simplifying the form of the equation.

3. How do you use an integrating factor to solve a differential equation?

To use an integrating factor, you first need to identify the differential equation as separable, linear, or exact. Then, you can use the formula for finding the integrating factor, which involves taking the exponential of an integral. Finally, you can multiply both sides of the equation by the integrating factor and solve the resulting equation.

4. What is the purpose of checking work when solving a differential equation using an integrating factor?

The purpose of checking work is to ensure that the solution obtained is correct. This involves substituting the solution back into the original differential equation and verifying that it satisfies the equation. It is important to check work to catch any mistakes that may have been made during the solving process.

5. Are there any special cases when using an integrating factor to solve a differential equation?

Yes, there are special cases when using an integrating factor. For example, if the integrating factor is equal to zero, then the differential equation becomes trivial and has no solution. Additionally, if the integrating factor is a constant, then it can be factored out of both sides of the equation, simplifying the solving process.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
158
  • Introductory Physics Homework Help
Replies
2
Views
734
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
589
  • Calculus and Beyond Homework Help
Replies
7
Views
623
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Electromagnetism
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
77
Back
Top