D.E. Linear equation with integrating factor

Click For Summary
The discussion revolves around solving a linear differential equation using an integrating factor. The user correctly identifies the integrating factor and derives the solution, but encounters issues with WeBWork rejecting their answer. After verifying the initial condition and substituting their solution back into the differential equation, they confirm the solution is correct. Ultimately, the problem was resolved by adjusting the notation to "exp(t)" for submission, highlighting the importance of adhering to specific formatting requirements in online homework systems.
Marco Lugo
Messages
9
Reaction score
0

Homework Statement



https://webwork.utpa.edu/webwork2_files/tmp/equations/2d/02a7e6a06f5b2424758fa01cc965f71.png with https://webwork.utpa.edu/webwork2_files/tmp/equations/80/81c176aa8964438a63eb096513245f1.png

Homework Equations


[/B]
Standard form: y' + p(x)y = f(x)

Integrating factor: r(x)= e∫p(x)dx

Solution: y(x) = e-∫p(x)dx(∫e∫p(x)dx ⋅ f(x) dx + C)

The Attempt at a Solution



From, y' - y = 7et + 56e8t

I got, p(x) = -1 and f(x) = 7et + 56e8t

So the integrating factor is r(t)= e∫-1 dt = e-t

Plug it into the solution formula,
y(t)= et [∫e-t (7et + 56e8t) dt + C]

= et (7t + 8e7t+ C)

= 7tet + 8e8t+ Cet

Using the initial condition y(0)= 2

= 8 + C = 2
C = -6

My final solution,
y(t) = 7tet + 8e8t- 6et

check: y(0) = 7(0)e(0) + 8e8(0)- 6e(0) = 2

But when I type it into WeBWork, it says my answer is incorrect. I don't think its a syntax problem because it has a preview button that allows you to see your actual problem.

Could it be that, y' - y = 7et + 56e8t is not in standard form?
 
Last edited by a moderator:
Physics news on Phys.org
Marco Lugo said:

Homework Statement



https://webwork.utpa.edu/webwork2_files/tmp/equations/2d/02a7e6a06f5b2424758fa01cc965f71.png with https://webwork.utpa.edu/webwork2_files/tmp/equations/80/81c176aa8964438a63eb096513245f1.png

Homework Equations


[/B]
Standard form: y' + p(x)y = f(x)

Integrating factor: r(x)= e∫p(x)dx

Solution: y(x) = e-∫p(x)dx(∫e∫p(x)dx ⋅ f(x) dx + C)

The Attempt at a Solution



From, y' - y = 7et + 56e8t

I got, p(x) = -1 and f(x) = 7et + 56e8t

So the integrating factor is r(t)= e∫-1 dt = e-t

Plug it into the solution formula,
y(t)= et [∫e-t (7et + 56e8t) dt + C]

= et (7t + 8e7t+ C)

= 7tet + 8e8t+ Cet

Using the initial condition y(0)= 2

= 8 + C = 2
C = -6

My final solution,
y(t) = 7tet + 8e8t- 6et

check: y(0) = 7(0)e(0) + 8e8(0)- 6e(0) = 2

But when I type it into WeBWork, it says my answer is incorrect. I don't think its a syntax problem because it has a preview button that allows you to see your actual problem.

Could it be that, y' - y = 7et + 56e8t is not in standard form?
I don't see anything wrong with your work or your solution. You can (and should) verify for yourself that your solution is correct by first checking that the initial condition is satisfied, and then by substituting your solution into the differential equation. Your solution should make the DE a true statement.

As far as why the software doesn't accept your solution, make sure that you are working the same problem it thinks you are working. I would also check with the instructor.
 
Last edited by a moderator:
Mark44 said:
I don't see anything wrong with your work or your solution. You can (and should) verify for yourself that your solution is correct by first checking that the initial condition is satisfied, and then by substituting your solution into the differential equation. Your solution should make the DE a true statement.

As far as why the software doesn't accept your solution, make sure that you are working the same problem it thinks you are working. I would also check with the instructor.

I checked the initial condition but you're right I should have substituted my solution into the problem.

my solution,
y(t) = 7tet + 8e8t - 6et

check:
dy/dt= 64e8t + 7tet + et

substituting both into the initial probelm,

dy/dt - y = 56e8t + 7et

64e8t + 7tet + et - 7tet - 8e8t + 6et

I get, 56e8t + 7et

I don't see anything wrong. I would submit the answer again but I only have 2 more tries, so I want to get some more advice before I do.
 
Marco Lugo said:
I checked the initial condition but you're right I should have substituted my solution into the problem.

my solution,
y(t) = 7tet + 8e8t - 6et

check:
dy/dt= 64e8t + 7tet + et

substituting both into the initial probelm,

dy/dt - y = 56e8t + 7et

64e8t + 7tet + et - 7tet - 8e8t + 6et

I get, 56e8t + 7et

I don't see anything wrong. I would submit the answer again but I only have 2 more tries, so I want to get some more advice before I do.

Your solution is correct.

Is there some special format you must adhere to when submitting solutions? For example, do you need to say exp(t) instead of et? Does it matter whether the terms come in a particular order, different from the one you wrote? For example, would ##8 e^{8t} + 7t e^{t} - 6e^{t}## or ##8 e^{8t} - 6 e^{t} + 7 t e^{t}## both be acceptable? Can you write ##t e^{t}## (or ##t \exp(t)##), or must it be in the form ##e^{t} \, t## (or ##\exp(t)\, t##), etc? I would certainly hope that the software would not be so picky, but who knows?
 
Ray Vickson said:
Your solution is correct.

Is there some special format you must adhere to when submitting solutions? For example, do you need to say exp(t) instead of et? Does it matter whether the terms come in a particular order, different from the one you wrote? For example, would ##8 e^{8t} + 7t e^{t} - 6e^{t}## or ##8 e^{8t} - 6 e^{t} + 7 t e^{t}## both be acceptable? Can you write ##t e^{t}## (or ##t \exp(t)##), or must it be in the form ##e^{t} \, t## (or ##\exp(t)\, t##), etc? I would certainly hope that the software would not be so picky, but who knows?

It worked! I had to use the exp(t) notation; which is weird because the software usually isn't picky and it would say if it wanted something expressed in a certain way.
Anyway thanks Mark44 and Ray Vickson for all you're help :D
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
7
Views
2K
Replies
7
Views
2K
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K