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7.7. THE RISCH DIFFERENTIAL EQUATION PROBLEM 191
7.7 The Risch Differential Equation Problem
In this section we solve case (b), n = 0 of Theorem 33: viz.
to find an algorithm which, given elements f and g of C(x) (with f
such that exp( ∫ f) is a genuinely new expression, i.e., for any nonzero
F with F ′ = fF , M = C(x, F ) is transcendental over C(x) and
has M const = C), finds y ∈ C(x) with y ′ + fy = g, or proves that
such a y does not exist. This is written as RDE(f, g).
These conditions on f mean that it is not a constant, and its integral is not
purely a sum of logarithms with rational number coefficients.
We will first consider the denominator of y ∈ C(x). We could assume that
C is algebraically closed, so that all polynomials factor into linears, but in fact
we need not. We will assume that we can factor polynomials, though we will
see afterwards that this is algorithmically unnecessary.
Let p be an irreducible polynomial. Let α be the largest integer such that
p α divides the denominator of y, which we can write as p α ‖ den(y). Let β and
γ be such that p β ‖ den(f) and p γ ‖ den(g). So we can calculate the powers of
p which divide the terms of the equation to be solved:
y ′
}{{}
α+1
There are then three possibilities.
+ fy
}{{}
α+β
= g
}{{}
γ
1. β > 1. In this case the terms in p α+β and p γ have to cancel, that is we
must have α = γ − β.
2. β < 1 (in other words, β = 0). In this case the terms in p α+1 and p γ must
cancel, that is, we must have α = γ − 1.
3. β = 1. In this case, it is possible that the terms on the left-hand side cancel
and that the power of p which divides the denominator of y ′ + fy is less
than α+1. If there is no cancellation, the result is indeed α = γ−1 = γ−β.
So let us suppose that there is a cancellation. We express f and y in partial
fractions with respect to p: f = F/p β + ˆf and y = Y/p α + ŷ, where the
powers of p which divide the denominators of ˆf and ŷ are at most β−1 = 0
and α − 1, and F and Y have degree less than that of p.
.
y ′ + fy = −αp′ Y
p α+1 + Y ′
p α + ŷ′ + F Y ˆfY
+
pα+1 p α
+ F ŷ
p + ˆfŷ. (7.47)
For there to be a cancellation in this equation, p must divide −αp ′ Y +F Y .
But p is irreducible and Y is of degree less than that of p, therefore p and
Y are relatively prime. This implies that p divides αp ′ − F . But p ′ and F
are of degree less than that of p, and the only polynomial of degree less
than that of p and divisible by p is zero. Therefore α = F/p ′ .