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I'm on my last homework problem, and I'm having some difficulty solving it:

$$2x(y+1)\ dx - y\ dy = 0, \quad y(0) = -2$$

I've gotten it into the form:

$$2x = \frac{yy'}{y+1}$$

but I don't know how to integrate the right-side. I'm not even sure what technique I would use.

Plugging the problem into Mathematica gives:

$$y(x) = -1 - W(e^{1 - x^{2}})$$

where W is the Lambert W function ... which I've never even heard of before, so I'm not sure how I'm expected to solve this using typical methods (I'm a chemical engineer).

  • 0
    $xe^x=y \iff W(y)=x$2017-02-19
  • 0
    From the formula $$f^{-1}(x)=\frac{1}{f'(f^{-1}(x))},$$ $$W'(x)=\frac{W(x)}{x+W(x)}.$$2017-02-19

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Integrate both sides of the equation $2x = \frac{yy'}{y+1}$ with respect to x to get: $$x^2/2 = y-ln(y+1)+C$$

To integrate $\frac{y(x)y'(x)}{y(x)+1}$ with respect to x make the substitution u=y(x)

  • 1
    Did you mean u = y(x) + 1?2012-11-15
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    Your substitution is easier. The substitution I gave will work too2012-11-15