Tough Calculation, novice mathematica user

Question

I have an equation, that I've been calling $b_N(x)$ that satisfies the following identity: $$-Nb_N(x)^2=(x-N)b_N(x)+xb_N'(x)$$ where $b_N'(x)$ is the first derivative. I take the derivative then of both sides and multiply both sides by $x$ giving me $$-2Nb_N(x)xb_N'(x)=xb_N(x)+\left[x^2+(1-N)x\right]b_N'(x)+x^2b_N''(x)$$ Using the original identity (but rewriting as $xb_N'(x)=(N-x)b_N(x)-Nb_N(x)^2$) and substituting on the LHS gives me $$-2Nb_N(x)\left[(N-x)b_N(x)-Nb_N(x)^2\right]=xb_N(x)+\left[x^2+(1-N)x\right]b_N'(x)+x^2b_N''(x)$$ Simplifying the LHS gives $$2N^2b_N(x)^3-2(x-N)(-Nb_N(x)^2)=xb_N(x)+\left[x^2+(1-N)x\right]b_N'(x)+x^2b_N''(x)$$ Finally, on the LHS, doing another substitution for $-Nb_N(x)^2$ with the original identity, moving the terms and simplifying gives me $$2N^2b_N(x)^3=\left[2x^2+(1-4N)x+2N^2\right]b_N(x)+\left[3x^2+(1-3N)x\right]b_N'(x)+x^2b_N''(x)$$

This calculation by hand is not very tedious and) I've done it a fair amount to verify. My interest though is to continue this process, (in other words, take the derivative of the last identity) and by hand it is really difficult and easy to make mistakes. For example, the derivative of both sides multiplied by $x$ is $$6N^2b_N(x)^2\cdot b_N'(x)=\left[4x^2+(1-4N)x\right]b_N(x)+...$$ I think you get the idea...So I would like to use Mathematica to tackle the problem. My problem is I'm not savvy with mathmatica. The function $b_N$ really is immaterial and it is the process of continued derivative taking and substitution that makes it hard. What is the best approach for handling this type of calculation in Mathematica?

Answer 1

Try this:

 eq = -n*b[x]^2 == (x - n)*b[x] + x*b'[x];
DSolve[eq, b[x], x]

The result is here:

{{b[x] -> (E^-x x^n)/(C[1] - n Gamma[n, x])}}

Here C[1] is the arbitrary constant.

Have fun!