MATH 3D Homework 6 Name: Solutions
Feb. 17, 2006
Section 2.2 (p. 140) Problems
8. Solve the initial-value problem
d2y
dt2−6dy
dt +y= 0; y(2) = 1, y0(2) = 1.
Solution: (10 points) This is a second-order, linear homogeneous problem
with constant coefficients. So, we look for solutions of the form y=ert:
r2−6r+ 1 = 0 =⇒r= 3 ±2√2.
Let us name r1= 3 −2√2 and r2= 3 + 2√2. Noting that the initial time is 2,
the general solution is
y(t) = c1er1(t−2) +c2er2(t−2).
So, by the initial conditions, we have
1 = y(2) = c1+c2
1 = y0(2) = r1c1+r2c2.
If we multiply the first equation by r1and subtract the second equation from
the first, then
(r1−r2)c2=r1−1 =⇒c2=r1−1
r1−r2
.
So,
c1= 1 −c2=1−r2
r1−r2
.
Plugging back in r1and r2, we can find y(t):
y(t) = µ1−r2
r1−r2¶e(3−2√2)(t−2) +µr1−1
r1−r2¶e(3+2√2)(t−2)
=Ã√2+1
2√2!e(3−2√2)(t−2) +Ã√2−1
2√2!e(3+2√2)(t−2)
=Ã√2+1
2√2!e−6+4√2e(3−2√2)t+Ã√2−1
2√2!e−6−4√2e(3+2√2)t.
Section 2.2 (p. 144) Problems
19. Find the general solution of
t2d2y
dt2+ 2tdy
dt + 2y= 0, t > 0.
Solution: (10 points) This is a second-order, linear homogeneous equation,
and it’s an Euler’s equation. So, we look for solutions of the form y=tr:
t2r(r−1)tr−2+ 2trtr−1+ 2tr=tr¡r2+r+ 2¢= 0.
Document URL: http://math.uci.edu/~pmacklin/Math3Dwinter2006.html
Date: February 22, 2006 Page 1 of 2
