MATH 3D Homework 8 Name: Solutions
Feb. 28, 2006
Section 2.4 Problems
9. Find two linearly independent solutions of t2y00 −2y= 0 of the form y(t) = tr.
Using these solutions, find the general solution of t2y00 −2y=t2.
Solution (12 points): The ODE is an Euler’s equation, so we seek solutions
of the form y=tr. Plugging this into the differential equation, we have
t2r(r−1)tr−2−2tr=tr(r2−r−2) = tr(r+ 1)(r−2) = 0.
Thus, two solutions are y1=t−1and y2=t2. We seek a a particular solution ψ
of the form
ψ=u1(t)y1(t) + u2(t)y2(t).
However, we note that the differential equation is not in the correct form y00 +
p(t)y0+q(t)y=g(t), so we first fix this:
y00 −2
t2y= 1.
We compute that
W[y1, y2] = det µy1y2
y0
1y0
2¶= det µt−1t2
−t−22t¶= 3.
Now, recall that
u0
1=−gy2
W=−t2
3=⇒u1=−1
9t3
u0
2=gy1
W=t−1
3=⇒u2=1
3ln t.
So, the general solution is given by
y(t) = c1y1+c2y2+ψ
=c1t−1+c2t2−1
9t3t−1+1
3ln tt2
=c1t−1+µc2−1
9¶t2+1
3t2ln t.
Notice that we could redefine ec2=c2−1
9.
Section 2.5 Problems
4. Find a particular solution of y00 +y0+y= 1 + t+t2.
Solution (13 points): The ODE is a constant coefficient problem, and the
righthand side is a polynomial. So, we judiciously guess that ψis a polynomial
pof some degree m. Let us determine the required degree.
Document URL: http://math.uci.edu/~pmacklin/Math3Dwinter2006.html
Date: February 28, 2006 Page 1 of 2
