Midterm 1 solutions / 2006.10.12 / Differential Equations II / MAT 3623.001
1. Use the definition to compute the Laplace transform of t e−2tu(t−3).
For which sdoes the transform converge?
Z∞
0
e−st t e−2tu(t−3) dt =Z∞
3
t e−(s+2)tdt ="−t e−(s+2)t
s+ 2 #∞
3
+Z∞
3
e−(s+2)t
s+ 2 dt
="−te−(s+2)t
s+ 2 −e−(s+2)t
(s+ 2)2#∞
3
="−e−(s+2)t
(s+ 2)2[t(s+ 2) −1]#t=∞
t=3
→e−3s−6(3s+ 5)
(s+ 2)2⇔s > −2
2. Find the inverse Laplace transform of ln(s−4).
Let F= ln(s−4). Since L[tnf] = (−1)ndnF
dsn, we have
L[tf] = −dF
ds =−1
s−4=L[−e4t], so tf =−e4t, so f=−e4t
t
3. Use the method of Laplace transforms to solve the initial value problem
x00 +x=u(t−3), x(0) = 1, x0(0) = 2
Take L:s2X−s−2 + X=−e−3s
s, so (s2+ 1)X=e−3s
s+s+ 2. Solve for X:
X=e−3s
s(s2+ 1) +s+ 2
s2+ 1 =e−3s1
s−s
s2+ 1+s
s2+ 1 +2
s2+ 1
Thus, x= [1 −cos(t−3)] u(t−3) + cos(t) + 2 sin(t)
4. Find the Taylor series about t= 0 of t5(4 + t2)−1. Use the summation notation, but also
write out the first three nonzero terms. What is the radius of convergence? Explain.
Let x=−t2/4. Then t2=−4x, so
t5
4 + t2=t5
4−4x=t5
4·1
1−x=t5
4
∞
X
k=0
xk=t5
4
∞
X
k=0 −t2
4!k
=
∞
X
k=0
(−1)k
4k+1 t2k+5 =1
4t5−1
16t7+1
64t9+...
The nearest singularities to the origin are 2iand −2i, so the radius of convergence is 2.
Alternately you can use the ratio test:
(−1)k+1t2k+7
4k+2 ·4k+1
(−1)kt2k+5
=|t|2
4<1⇔ |t|<2
5. Find the first three nonzero terms of the power series solution about t= 0 to the initial
value problem
(t+ 1)x00 −x= 0, x(0) = 0, x0(0) = 2
Let x=
∞
X
k=0
aktk, where a0= 0, a1= 2. Then x00 =
∞
X
k=0
ak+2(k+ 2)(k+ 1)tk,
so tx00 =
∞
X
k=0
ak+2(k+ 2)(k+ 1)tk+1 =
∞
X
k=1
ak+1(k+ 1)ktk=
∞
X
k=0
ak+1(k+ 1)ktk
Plug into tx00 +x00 −x= 0, collect the coefficients of xkto obtain the recurrence relation
ak+1(k+ 1)k+ak+2(k+ 2)(k+ 1) −ak= 0, and solve:
ak+2 =ak−ak+1(k+ 1)k
(k+ 2)(k+ 1) ,i.e. ak=ak−2−ak−1(k−1)(k−2)
k(k−1)
Choosing k= 2,3, ... we can obtain further coefficients: a2=a0−a1·1·0
2·1= 0,
a3=a1−a2·2·1
3·2=1
3,a4=a2−a3·3·2
4·3=−1
6, so x= 2t+1
3t3−1
6t4...
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