Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    7
    ReputationRep:
    Solved it for the first initial conditions, but I would appreciate a check.

    What is the situation with the underlined new initial condition though?
    Attached Images
       
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    Solved it for the first initial conditions, but I would appreciate a check.

    What is the situation with the underlined new initial condition though?
    I presume it's just the limit criterion that's causing the problem.

    \displaystyle \lim_{x\to\infty} y(x)=\lim_{x\to\infty}(Ae^{2x}+B  e^{-x})=0

    So,

    \displaystyle 0 =\lim_{x\to\infty}(Ae^{2x})+\lim  _{x\to\infty}(Be^{-x})

    What's the limit for the second term?

    Then what must A be for the whole thing to equal 0?

    Edit: First one looks fine, though to finish off, I would restate y(x)= ... with the values of A,B filled in.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    I presume it's just the limit criterion that's causing the problem.

    \displaystyle \lim_{x\to\infty} y(x)=\lim_{x\to\infty}(Ae^{2x}+B  e^{-x})=0

    So,

    \displaystyle 0 =\lim_{x\to\infty}(Ae^{2x})+\lim  _{x\to\infty}(Be^{-x})

    What's the limit for the second term?

    Then what must A be for the whole thing to equal 0?

    Edit: First one looks fine, though to finish off, I would restate y(x)= ... with the values of A,B filled in.
    Isn't the limit for the second term 0 though? Subbing in a large value, I get back 0.
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    Isn't the limit for the second term 0 though? Subbing in a large value, I get back 0.
    Yes. You can also write it as 1/e^x. So, as x goes to infinity, e^-x goes to zero.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Yes. You can also write it as 1/e^x. So, as x goes to infinity, e^-x goes to zero.
    OK. But the other term is infinity, no? So isn't that giving that 0 = infinity?
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    OK. But the other term is infinity, no? So isn't that giving that 0 = infinity?
    Well, there's one particular value of A for which it doesn't go to infinity.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Well, there's one particular value of A for which it doesn't go to infinity.
    0?

    Wait, A and B are both 0?
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    0?
    Yes, so A=0

    Wait, A and B are both 0?
    No - why would you think that?
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Yes, so A=0



    No - why would you think that?
    LHS = 0

    Therefore, RHS = 0

    If the first term when e^(2x) approaches infinity is infinity, I need A = 0, as nothing else gives a non-positive result (that I know).

    So the e^(-x) is hugely negative when infinity is approached, but I need B to make it 0.

    So I incorrectly assumed B = 0.
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    LHS = 0

    Therefore, RHS = 0

    If the first term when e^(2x) approaches infinity is infinity, I need A = 0, as nothing else gives a non-positive result (that I know).
    Yes.

    So the e^(-x) is hugely negative when infinity is approached, but I need B to make it 0.
    No. You've already covered this. See posts 3 and 4.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Yes.



    No. You've already covered this. See posts 3 and 4.
    OK, so the e term will always give 0. Does that mean B could take any value?
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    OK, so the e term will always give 0. Does that mean B could take any value?
    Based on the limit criterion, yes, B could by anything. But you also have another condition, and you can use that to now determine B.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Based on the limit criterion, yes, B could by anything. But you also have another condition, and you can use that to now determine B.
    My other condition is that y(0) = 1 .

    y(x) = Ae^(2x) + Be^(-x)

    So...

    1 = Ae^(2*0) + Be^(-0)

    1 = A + B

    B = 1 - A

    If A is 0, then B = 1
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    My other condition is that y(0) = 1 .

    y(x) = Ae^(2x) + Be^(-x)

    So...

    1 = Ae^(2*0) + Be^(-0)

    1 = A + B

    B = 1 - A

    If A is 0, then B = 1
    Yep. Done.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Yep. Done.
    One last question on this problem. It would be written in its final form as...

    y(x) = e^(-x)

    Having subbed in A = 0 and B = 1
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    One last question on this problem. It would be written in its final form as...

    y(x) = e^(-x)

    Having subbed in A = 0 and B = 1
    Yes.
    • Thread Starter
    Offline

    7
    ReputationRep:
    (Original post by ghostwalker)
    Yes.
    Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
    • Study Helper
    Offline

    13
    (Original post by ChrisLorient)
    Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
    Looking at the RHS of your DE, if there was a PI, then it's clearly 0, hence none.

    Edit: That is treating the PI as something separate to the complementary functions.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.