1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted