# Having trouble solving a differential equation: 2x(y+1)dx - ydy = 0

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).

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## 1 Answer

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)

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Oh wow, that's pretty simple actually. Thanks! – Nick Nov 15 '12 at 11:31
Did you mean u = y(x) + 1? – Nick Nov 15 '12 at 11:35
Your substitution is easier. The substitution I gave will work too – Amr Nov 15 '12 at 11:36