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General solution of differential equation

  1. Oct 16, 2007 #1
    1. The question asks me to show that e^x is a solution of xy'' - (2x+1)y' + (x+1)y=0 and find the general solution.



    2. I managed to simplify the equation to u''xe^(x) - u'e^(x) = 0 by letting y=ue^(x) and finding the differentials and substituting them in.
    I've then let z=u dz/du=u' and d^2z/du^2 = u''
    so I get xe^(x)(d^2z/du^2) - e^(x)dz/du = 0
    How would I solve this?
     
  2. jcsd
  3. Oct 16, 2007 #2
    use integration by parts
     
  4. Oct 16, 2007 #3
    Sorry but I am still confused;

    if xe^(x)(d^2z/du^2) - e^(x)dz/du = 0
    can I simplify this to

    x(d^2z/du^2) = dz/du

    xdz = du
    xz = u +c but z = u???
    xu = u +c
    How does this help find the general solution?
     
  5. Oct 17, 2007 #4

    HallsofIvy

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    You don't solve that! It has two independent variables, u and x. Surely, you don't mean "let z=u dz/du=u' and d^2z/du^2 = u''. If you let z= u, then dz/du= 1. Perhaps you meant "let z= u, dz/dx= u'". But in that case you've gained nothing- you've just renamed u. Much better is "let z= du/dx, dz/dx= d2u/dx2". Then your equation becomes the first order equation xe^(x) z'- e^(x) z= 0. I would be inclined to first divide the entire equation by e^(x)!
     
  6. Apr 30, 2011 #5
    Hi can someone please help me with: general solution of dy/dx - y = x + 2x^2

    i know how to find general solutions but only when i can seperate the y and x in 2 sides and multiply with dx and dy. someone please help me. i have looked in many books
     
  7. Apr 30, 2011 #6

    HallsofIvy

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    It is NOT good idea to add on to someone elses's thread. People who have already responded to the thread may not even look at your post. Use the "new thread" button on the main menu. In this case, I have already responded on the thread you did start about 4 minutes later!
     
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