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Find an explicit solution of the given initial-value problem

  1. Sep 26, 2014 #1
    Hello all, hoping to get a bit of help with a diff eq problem that goes like this:

    Find an explicit solution of the given initial-value problem,

    x^2*(dy/dx)=y-xy ; y(-1) = -6

    I separated my variables to get:

    S (1-x)/x^2 dx = S 1/y dy

    Integration of left via partial fractions:

    S 1/x^2 dx - S 1/x dx = -1/x - ln|x|

    Integration of right :

    S 1/y dy = ln|y|

    therefore :

    -1/x - ln|x| = ln|y| + C

    Now this is the first Calc class I have taken in a while so I am assuming more than one error here:

    e^(-1/x) -x = y + e^C
    y = e^(-1/x) -e^C - x
    y = e^(1/x - C) -x

    Did I solve for y correctly here? If not is my issue in the integration? Also at what point is it best to solve for C in this equation?

    Thank you in advance for any assistance.
     
  2. jcsd
  3. Sep 27, 2014 #2

    Char. Limit

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    Gold Member

    You had it right up to here. That's definitely good.

    You're kinda applying the exponential operator all wrong here. Your first step should look like

    [tex]e^{-\frac{1}{x} - ln|x|} = e^{ln|y| + C}[/tex]

    From there, you'll want to separate the terms on the left via the exponential rule:

    [tex]e^{a+b} = e^a e^b[/tex]

    You should get something akin to [itex]e^{-1/x} \times \frac{1}{x}[/itex] on the left side, and of course [itex]e^C y[/itex] on the right side. From there, it should be rather easy to simplify and plug in your initial condition to solve for e^C.
     
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