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DiffEq - Initial Value Problem / Integration help

  1. Jun 19, 2007 #1
    This question is an initial value problem for diffeq. We are asked to solve explicitly for y.

    (1+cos(x))dy = ((e^(-y))+1)*sin(x)dx , y(0) = 0

    I attempted a separation of variables and ended up with the following:

    dy / ((e^(-y))+1) = (sin(x) / (1 + cos(x))) dx

    I know that my next step is to integrate both sides and then solve using the given initial value, but I am unsure as to how I am supposed to integrate either side.

    For the right side, I believe I can integrate using u-substitution, where:

    u = cos(x) + 1
    du / dx = -sin(x)

    So that the right side becomes -1/u, integrates to -ln(u), then -ln(1+cos(x)).

    For the left side, I've tried using partial fraction decomposition but end up either with my original equation or the natural logarithm of a negative number.
    This is where I need help, is in the integration of the left hand side.

    thank you
  2. jcsd
  3. Jun 19, 2007 #2
    IF you mutiply dy/((e^-y)+1) by (e^y)/(e^y) before intetgrating, can you then use integration by substitution?
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