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

    What is the situation with the underlined new initial condition though?
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    (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.
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    (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.
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    (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.
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    (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?
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    (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.
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    (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?
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    (Original post by ChrisLorient)
    0?
    Yes, so A=0

    Wait, A and B are both 0?
    No - why would you think that?
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    (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.
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    (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.
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    (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?
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    (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.
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    (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
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    (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.
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    (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
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    (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.
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    (Original post by ghostwalker)
    Yes.
    Thinking about this again, why have I not found a particular integral? Is it because it's homogenous?
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    (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.
 
 
 
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