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D.E.: Boundary Value Problem

  1. Feb 25, 2013 #1
    I'm not sure if my answer is correct. Did I make a mistake somewhere? I'm not sure the ± needs to be there.
  2. jcsd
  3. Feb 25, 2013 #2


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    How in the world did you get a quadratic equation out of this? [itex]y(2)= (C_1+ 6C_2)e^6= e^6[/itex] and [itex]y'(1)= (3C_1+ 4C_2)e^3= e^3[/itex]. The derivative is [itex]y'= 3C_1e^{3x}+ C_2e^{3x}+ 3C_2xe^{3x}= ([3C_1+ C_2]+ 3C_2x)e^x[/itex]. It does not involve "[itex]C_1C_2[/itex]"!

    You have [itex]C_1+ 6C_2= 1[/itex] and [itex]3C_1+ 4C_2= 1[/itex], two linear equations.
  4. Feb 25, 2013 #3
    c1 is represented by c, c2 is represented by d


    That's y'

    I did make an error by leaving out the + sign between c1 and c2 for y'

    That makes c1 = -1 and c2 = 1
    Last edited: Feb 25, 2013
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