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Diff EQ- exponential raised to the exponential?

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the equation:

    [tex](x+xy^{2})dx + e^{x^{2}}y dy =0[/tex]

    2. Relevant equations

    n/a

    3. The attempt at a solution

    i) [tex]xdx(1+y^{2}) = -e^{x^{2}}y dy[/tex]
    (multiply both sides by 2 to prepare for integration:

    ii) [tex]\frac{-2xdx}{e^{x^{2}}}= \frac{2y dy}{1+y^{2}}[/tex]

    integrate and get:
    iii) [tex]\frac{1}{e^{x^{2}}}= ln(1+y^{2}) + C[/tex]

    first integral was acheived from Maple, im not sure if its what should be used here, but its what i have for now

    iv) exponentiate both sides to remove natural log:

    [tex]e^{e^{-x^{2}}}=1 + y^{2} + C[/tex]

    v)combine constants, separate y

    [tex] y^{2} = e^{e^{-x^{2}}} - C [/tex]

    vi)square root of the system:

    [tex]y=\sqrt{ e^{e^{-x^{2}}}} -C[/tex]


    Does this seem like a reasonable answer? I'm wary because of the occurance of the exponential function, raised to the exponential function.
     
  2. jcsd
  3. Feb 11, 2009 #2

    Mark44

    Staff: Mentor

    One thing that's nice about differential equations is that when you get a solution, you can check it by seeing if it satisfies the DE.

    As a side note, you show the same C in your last three equations. Actually these are all different constants, plus the third equation does not follow from the second. sqrt(a + b) != sqrt(a) + sqrt(b).
     
  4. Feb 11, 2009 #3
    yea, im still getting used to keeping track of my C's...thanks for that note

    i dont know why i didnt think to substitute the solution back in. thanks!
     
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