Results 1 to 10 of 10

Math Help - Solve Initial Value Problem given General Solution.

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    48

    Solve Initial Value Problem given General Solution.

    Im having a little trouble with these problems. Im sure its pretty simple. Heres one from my book:


    Given the family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

     y = C1e^x + C2e^-^x, (-\infty, \infty)
     y'' - y = 0, y(0) = 0, y'(0) = 1


    Can someone show me how to do this problem? I am just confused one how to work it. Im sure its pretty simple but I cant seem to figure it out.

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Your goal is to plug in the initial conditions y(0)=0, y'(0)=1, and solve for C_{1},C_{2}. So what two equations do you get by plugging in the initial conditions into your general solution?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    48
    plugging in  y(0) = 0
    C1 + C2 = 0
    plugging in  y'(0) =  1
    C1 - C2 = 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Good so far! What next?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    48
    add the equations together?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Sure. What does that give you?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    Posts
    48
    C1 = 1/2
    C2 = -1/2


    Then
     y = .5e^x - .5e^-^x

    thats it right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    This is the beautiful thing about differential equations: it's quite straight-forward to check your own answer (hardly more difficult than differentiating, and quite a bit easier than integrating). So I'm going to ping this question right back at you:

    1. Does your solution satisfy the DE?
    2. Does your solution satisfy the initial conditions?

    If so, I'd say you've got it.

    You should get into the habit of doing two things to every solution to every DE you ever solve: 1. Find the interval of validity. 2. Check your solution by differentiation.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Oct 2009
    Posts
    48
    Thank you!

    the reason I wasnt sure what quite to do is that my teacher crammed 3 sections into an hour since we are behind. Thanks for the help tho, i appreciate it!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're welcome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Existence of a solution; Initial value problem
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: December 3rd 2010, 11:35 PM
  2. series solution of an initial value problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 27th 2010, 09:52 AM
  3. solve given general solution
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: January 13th 2010, 02:35 AM
  4. Initial Value Problem General Question
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 14th 2009, 08:05 AM
  5. Replies: 3
    Last Post: April 23rd 2007, 06:08 PM

Search Tags


/mathhelpforum @mathhelpforum