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I'm solving the below differential equation:

$$ 3z^2 U'(z) -2 U(z)^2 - z U(z) + 2 = 0$$

I have two boundary conditions - $U(0)=1$, $U'(0) = -\frac{1}{4}$, however it is apparent that the set of solutions to this equation, before imposing the initial conditions, satisfy the first initial condition $U(0)=1$. Looking at the differential equation this is because it can be rearranged to:

$$ [U(0)]^2 = 1 $$

So it is evident that all of the general solutions should fit this condition. What's more, through "numerical experiment", I believe that the first derivative condition should be satisfied as well (though I can't prove it from the differential equation).

I am not sure that I can set the constant of integration to be anything I want, as this changes the behaviour of the solution. However, I can simplify the solution greatly by choosing "convenient" values of this constant.

Also the resulting function is a function of $\frac{1}{z}$, so it cannot actually be evaluated at 0, so I have been using limits, which feels quite mathematically dodgy.

This is a peculiarity that I have not come up against before - how exactly am I to interpret this?

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    Is $z$ a parameter or the independent variable?2011-12-22
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    Independant variable, sorry. It's not typical notation, so I'll make it more clear.2011-12-22
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    The real issue is that $3z^2=0$ when $z=0$, so you cannot apply the fundamental existence/uniqueness theorem. In fact, your calculation (forcing $U(0)^2=1$) shows that this equation has **no** solution if we try to impose an initial condition where $U(0)^2\not=1$.2011-12-22

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