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Homework Help: Second order DE problem with initial condition rather than 0

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the following DE

    y'' + 8y' − 9y = 0, y(1) = 1, y'(1) = 0

    2. Relevant equations

    Homogenous DE with constant coefficients

    3. The attempt at a solution

    Well, i solved it normally using a CE and having
    yH= c1 e^t + c2 e^(-9t) ..
    y' = c1 e^t -9 c2 e^(-9t)

    I then plugged in 1 in each of the above equation set from the homo. equation and its derivative and set that

    c1 e^1 + c2 e^-9 = 1
    c1e^1 -9c2 e^-9=0

    I solved the eqns getting c1 =0.33 and c2= 810 getting my final homo. eqn as

    y= 0.33 e^t + 810 e^ -9t

    HOWEVER, this isn't the answer. The forum asks for my attempt NO? ... Well, the detailed answer just confuses me... Te fact that the I.C is y(0) and it is y(1) instead changes the procedure. I don't know why.. The book gives yh= k1 e^(t-1) + k2 e^-9(t-1) giving a note that c1=k1 e^-1 and c2= k2 e^9 ... They then write y'= k1 e^(t-1) -9 k2 e^-9(t-1)...

    I am sorry for making this long but can any PATIENT person give me an explanation to this. I am pretty confused here and thanks in advance to whoever shows up with a good reply. I appreciate it a LOT! :D
  2. jcsd
  3. Nov 18, 2013 #2


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    Education Advisor

    No procedure has been changed. From looking at what you post, it appears as if the book choose to use properties of exponents to simplify their anticipated answer.
  4. Nov 18, 2013 #3


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    Homework Helper
    Gold Member

    Also, all you have to do is check whether, for your answer, ##y(1) =1## and ##y'(1) = 0## to within reasonable errors, since you truncated the constants.
  5. Nov 18, 2013 #4
    can someone kindly teach me the 'properties of exponents to simplify their anticipated answer.' in this context???
  6. Nov 18, 2013 #5

    D H

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    Staff Emeritus
    Science Advisor

    You're correct. That isn't the answer. It is close to the answer. The issue is that you solved for c1 and c2 numerically. Why? It wouldn't have taken any more effort to solve for it symbolically and obtained an exact answer.

    Two things you should notice here. One is that this shift simplifies things a bit. The other is that this is exactly the same as your answer had you solved for your c1 and c2 exactly.
  7. Nov 20, 2013 #6
    Thanks a lot guys for your kind hands! I just have a small part falling out. The SHIFT! Consider we begin at y(0) and the roots of the CE is 1 and -1 .. so the homo. equation will be y= c1 e ^( t ) + c2 e^(-t).. Now consider when it begins at 2.. How to do the shift or what steps were done to write the new SHIFTED equation as y=k1 e^ (t-2) + c2 e^-1(t-2)??? Thanks again for your patience
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