Answer Key for Quiz 8 (section B)
1. If y=ax6+bx5+cx, then
dy
dx = 6ax5+ 5bx4+cand d2y
dx2= 30ax4+ 20bx3.
Plugging all this into the differential equation x2d2y
dx2−5xdy
dx + 5y=x6we get
x6=x2¡30ax4+ 20bx3¢−5x¡6ax5+ 5bx4+c¢+ 5 ¡ax6+bx5+cx¢
= 30ax6+ 20bx5−30ax6−25bx5−5cx + 5ax6+ 5bx5+ 5cx
= 5ax6.
Since this is supposed to hold for all x, we must have a=1
5. There are no restrictions on band csince all
those terms cancelled. In fact, every solution of this differential equation has the form y=1
5x6+bx5+cx
for some constants band c.
2. Since the density depends on the distance from the ocean, we have to slice the city along “streets” parallel
to the ocean. A street xmiles from the ocean has length 2(6 −x) by similar triangles.
If the street has thickness dx, then the number of people living on it is 20000(6 −x)e−xdx. The distance
from the ocean is between 0 and 6, so the total population of the city is
P= 20000 Z6
0
(6 −x)e−xdx.
To evaluate this integrate by parts with dv =e−xdx and u= 20000(6 −x), so that du =−20000 dx and
v=−e−xand we have
P=−20000(6 −x)e−x¯
¯
6
0−20000 Z6
0
e−xdx
= 0 − −20000(6)(1) + 20000e−x¯
¯
6
0
= 120000 + 20000e−6−20000
= 100000 + 20000e−6
≈100049.575
≈100050.