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As part of my solution to a problem, I come to a point where I need to find the solutions to $-2\partial_{T}B\left(T\right)+\frac{3}{4}B\left(T\right)\left(A\left(T\right)^{2}+B\left(T\right)^{2}\right)=0$

$2\partial_{T}A\left(T\right)+\frac{3}{4}A\left(T\right)\left(B\left(T\right)^{2}+A\left(T\right)^{2}\right)=0$

where $\partial_{T}(f)$ is the derivative with respect to $T$.

It is possible that I made a mistake in the steps leading to this because I am supposed to be able to get a not-so-ugly solution for $A(T)$ and $B(T)$. Is there one that exists and I don't see it? I've tried the following: alt text

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You can make the second terms in both equations vanish by multiplying the first by $A(T)$, the second by $B(T)$, and subtracting. The resulting equation is readily solved for the product $A(T)B(T)$, reducing the system to a single ODE which is directly integrable.

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I hope one day to exploit the symmetry more intuitively. Thanks! Unfortunately, I had a sign error in the TeX math. The signs should be as they are in the firs two lines of the image. Can I still solve something that looks like d(A(T)*B(T) = 0 ? –  Gus Oct 18 '10 at 15:26
    
@Gus: Yes, you're close. (BTW, I wonder why the first line in the image uses two minus signs. What's the point? Is there perhaps a typo in it?) When you see an expression like a'b - b'a in an ODE, be reminded of the quotient rule and consider a substitution of the form f = a/b or f = b/a. You should check the possibility a = b = 0 as a special case, for then neither fraction is defined. –  whuber Oct 18 '10 at 15:38

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