Math 3113, Quiz IV, Solution
1. Consider the initial value problem
tx000 + 2x00 + (t−1)x0= 0 x(0) = x0(0) = x00(0) = 0.
If L{x(t)}=X(s) then obtain the first order differential equation satisfied by X(s).
Write it in the separable form but DO NOT solve it.
We have L{tx000}=−(s3X(s))0=−(3s2X(s)+ s3X0(s)),L{2x00}= 2s2X(s) and L{(t−
1)x0}=L{tx0} − L{x0}=−(sX(s))0−sX (s) = −(X(s) + sX0(s)) −sX(s). Hence
applying Laplace Transform to the differential equation and collecting terms involving
X(s) and X0(s) we get
X0(s)(−s3−s) + X(s)(−s2−s−1) = 0 ⇒X0(s)
X(s)=−s2+s+ 1
s3+s
(Note that if you solve the above separable DE you get that X(s) equals the expression
given in problem 2 below. Hence the 2 problems are parts of just one problem.)
2. Find the Laplace inverse of
X(s) = 1
s(s2+ 1)
There are several ways to solve this :
(a) Using Convolution product Theorem :
L−1{1
s(s2+ 1)}=L−1{1
s2+ 1}∗L−1{1
s}= sin t∗1 =
t
Z
0
sin τdτ = 1 −cos t.
(b) Using Partial Fractions :
1
s(s2+ 1) =A
s+Bs +C
s2+ 1 =1
s−s
s2+ 1 ⇒ L−1{X(s)}= 1 −cos t.
(c) Using Transform formula for integrals :
L−1{1
s(s2+ 1)}=L−1{
1
s2+1
s}=
t
Z
0
L−1{1
s2+ 1}dτ =
t
Z
0
sin τdτ = 1 −cos t.
Hence, the solution is
x(t) = L−1{X(s)}= 1 −cos t.
1