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Homework Help: Bernoulli Equation

  1. Sep 19, 2010 #1
    I have solved the following Bernoulli equation by letting Z = y1 - 2:

    xy' - 2y = x3y2. I obtained the solution

    y = 5/(5c1 - x5)

    which Wolfram Alpha has confirmed.

    From this result, I have obtained y' to be

    y' = 25x4/(5c1 - x5)2

    The problem is when I go to check the solution by plugging into DE:

    x*25x4/(5c1 - x5)2 - 2*5/(5c1 - x5)

    [tex] = \frac{25x^4}{(5c_1 - x^5)^2} - \frac{10}{5c_1 - x^5} = \frac{25x^4}{(5c_1 - x^5)^2} - \frac{10*(5c_1 - x^5)}{(5c_1 - x^5)^2}[/tex]

    which will never equal the right hand side of the original equation:

    [tex]x^3y^2 = x^3\frac{25}{(5c_1 - x^5)^2}[/tex]


    Anyone seeing where I am messing this up?

    thanks.
     
  2. jcsd
  3. Sep 19, 2010 #2
    Hello fellow engineer ,

    First of all your y(x) and wolframs y(x) are not the same function.

    You are missing an [tex] x^{2}[/tex] term.
     
  4. Sep 19, 2010 #3
    Ahhh....yes. Thanks! So let's try this again....

    The solution to xy' - 2y = x3y2 is given by

    [tex]y(x) = \frac{5x^2}{5c - x^5} = 5x^2(5c - x^5)^{-1}[/tex]

    [tex]\Rightarrow y'(x) = (5x^2)\cdot[(-1)(5c - x^5)^{-2}(-5x^4)] + ((5c - x^5)^{-1})\cdot[10x][/tex]

    [tex]\Rightarrow y'(x) = \frac{25x^6}{(5c-x^5)^2}+\frac{10x}{5c-x^5}[/tex]

    [tex]\Rightarrow y'(x) = \frac{25x^6+(5c - x^5)(10x)}{(5c - x^5)^2}[/tex]

    Plugging the result back into original DE we have

    [tex]\frac{25x^7+(5c - x^5)(10x^2)}{(5c - x^5)^2} - 2*\frac{5x^2}{5c - x^5}=
    \frac{25x^7+(5c - x^5)(10x^2)}{(5c - x^5)^2} -\frac{(5c - x^5)(10x^2)}{(5c - x^5)^2} = \frac{25x^7}{(5c - x^5)^2} = A
    [/tex]

    From the right side of original DE we know that

    [tex]A' = x^3y^2 = \frac{(x^3)(25*x^4)}{(5c - x^5)^2} =\frac{25x^7}{(5c - x^5)^2} [/tex]

    Thus, A = A' and we must have the solution.

    Thanks for catching that error! :smile:
     
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