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2nd Order Homogeneous, Real Roots, Initial Value

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem

    question.jpg

    2. Relevant equations

    Quadratic Formula



    3. The attempt at a solution

    work.jpg

    My problem is that I don't understand how to solve the constants now, I understand, 2 equations, 2 unknowns, but when I plug the y(0) = 0 into the YsubH equation, everything cancels out, I can't solve for anything. I don't see anyway to cancel anything in the equations either to make it not equivalent to 0, or to be able to solve it.

     
  2. jcsd
  3. Jul 6, 2015 #2

    Mark44

    Staff: Mentor

    For simple differential equations such as this, it's better to type the equation and initial condition than to post an image. What I see is a thumbnail that shows "0, y". I have to click the image to see the full image.

    Your equation is
    2y'' + y' - 4y = 0, y(0) = 0, y'(0) = 1
    Your solution looks fine to me. Your error is in how you evaluated it. Note that e0 = 1, so you get ##c_1 + c_2 = 0## for your first equation, so everything does not cancel out.
     
  4. Jul 6, 2015 #3
    Okay, I'll remember that, the next time that I post... and thank you so much for catching my error, I feel terrible that I made such an algebra mistake...
     
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