Finding C with inital condition, its wrong, any ideas why?

In summary, the conversation involves solving for the value of c in the function y = ce^{-2x} + e^{-x} in order to satisfy the initial condition y(3) = 8. After plugging in values and trying different approaches, it is determined that the correct value of c is (8 - exp(-3))/exp(-6).
  • #1
mr_coffee
1,629
1
Hello everyone, I'm having troubles figuring out this problem. The directions say:
It is easy to check that for any value of c, the function
y = ce^{-2x} + e^{-x}
is solution of equation
y' + 2y = e^{-x}.
Find the value of c for which the solution satisfies the initial condition y(3)= 8.


So I pluged in 4 for x and y(x) i plugged in 8 and got:
8 = ce^(-2*3) + e^-(3)
8 = ce^-6 + e^-3
8/(e^-6+e^-3) = c
c = 153 which is wrong
i also tried ln(153) = 5, which is also wrong. any idea where i screwed up, thanks!
 
Physics news on Phys.org
  • #2
mr_coffee said:
...
y = ce^{-2x} + e^{-x}
...
Find the value of c for which the solution satisfies the initial condition y(3)= 8.
...
8 = ce^-6 + e^-3
8/(e^-6+e^-3) = c

ah, pesky algebra. 8 = c exp(-6) + exp(-3)
gives c = (8 - exp(-3))/exp(-6).
 
  • #3
Thanks a lot qbert, i don't know how i didn't see that hah! too much coffee tonight!
 

1. What is "C" and why is it important in finding initial conditions?

"C" represents the constant of integration in a differential equation. It is important because it allows us to find the specific solution that satisfies the given initial conditions.

2. Why does my solution for finding C with initial condition seem incorrect?

There could be several reasons for this. First, check your calculations to make sure there are no errors. Also, check if your initial conditions are correct and if they are consistent with the given equation. Another possible reason could be that the equation itself may not have a unique solution for the given initial conditions.

3. Can I use any value for "C" when finding initial conditions?

Yes, you can use any value for "C" as long as it satisfies the given initial conditions. However, it is important to note that a specific value of "C" may result in a simpler or more complex solution, so it is best to choose a value that makes the solution more manageable.

4. Why is it important to solve for "C" in finding initial conditions?

Solving for "C" allows us to find the specific solution that satisfies the given initial conditions. Without solving for "C", we would only have a general solution which may not accurately represent the system we are trying to model.

5. Are there any alternative methods for finding initial conditions without using "C"?

No, "C" is an essential part of finding initial conditions in a differential equation. However, there may be alternative methods for solving the equation that can simplify the process of finding "C". It is important to explore different methods and choose the one that best suits the given problem.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
335
  • Calculus and Beyond Homework Help
Replies
2
Views
280
  • Calculus and Beyond Homework Help
Replies
3
Views
281
  • Calculus and Beyond Homework Help
Replies
2
Views
898
  • Calculus and Beyond Homework Help
Replies
7
Views
501
  • Calculus and Beyond Homework Help
Replies
5
Views
291
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
714
Back
Top