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(2+x^2)dy/dx - x(y+3)= 0 , y(0)=-1

  1. Oct 22, 2005 #1
    The question asked to solve this initial value problem

    (2+x^2)dy/dx = x(y+3)

    (2+x^2)dy/(dx*(y+3) = x

    dy/[(y+3)*dx]= x/(2+x^2)

    dy/(y+3) = x*dx/(2+x^2)

    S dy/(y+3) = S x/(2+x^2)*dx

    ln(y+3) = ?

    Im not sure what to do with the RHS do I do integration by parts?
  2. jcsd
  3. Oct 22, 2005 #2


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    Staff Emeritus
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    Did it occur to you that, since you have xdx in the numerator, you might try a substitution: u= 2+ x2?
  4. Aug 13, 2008 #3
    well your right so far

    so your only problem is the rhs?
    if so
    let u=x^2+2

    the sub back into rhs as u=x^2+2 and dx=du/(2x)to be left with

    x's cancel
    then intergrate with respect to u, once done that sub u=x^2+2
    then i'm sure you can find the constant by using y(0)=-1

    only way to know how to do this is to practise many examples!
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