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270 Answers to Exercises and Solutions to Problems

and

p(t) = t”/9 - rt2/6 + c.

Substitute t = 0 into (1) and (3) to obtain 3c = c2 - rc, whence c = 0 or

c= r+3, and

(2~13 - r/3)(-r/3) = (-r/3)(r2/9),

whence 6c = 3r + r2. Hence (r,c) = (6,9) or (-3,0). The first leads to

p(t) = t3/9 - t2 + 9 which is valid, and the second to p(t) = t3/9 + t2/2

which is not (check t = 6). Hence p(t) = t3/9 or p(t) = t3/9 - t2 + 9.

In general, if p(t) = n ‘-‘Y, then p(p’(t)) = p’(p(t)).

5.10. The result holds for n = 1. Suppose it holds for n = m > 1. Then

(x + yp+') = (x + y)(x + y - 1p

(x - l)Wy(m-k)

= ; ) x(k)y(m+l-k)

= g ( k’c 1 ) xWy(m+l-k) + 2 ( ; ) x(~)~++~-~)

k=O

= .(~+lJ+$ [( k:l ) + ( y )] x@)Y(m+l--k)

+ y(“+l),

which yields the result for n = m + 1. The result follows by induction.

5.11. If y is a polynomial in x, the degree of the left side equals the degree
of y, so that y must be a cubic. Successive differentiations of the equation
yield
9y”’ + 4y” + d = 3x2 + 10x - 2
4~“’ + y” = 6x + 10
y”’ = 6.

Working up from the last equation, we find that y” = 6x-14, y’ = 3x2-14x
and y = x3 - 7x2 + 4. It is readily checked that this indeed satisfies the
differential equation.

5.12. f(x) has the form ax5 + bx4 + cx3. For some polynomial g(x), ax5 +
bx4 + cx3 - 1 = (x - 1)3g(x). D’ff 1 erentiating the equation twice and setting