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Quiz 4 Solutions - Introduction to Ordinary Differential Equations | MATH 3113, Quizzes of Mathematics

Material Type: Quiz; Class: Introduction to Ordinary Differential Equations; Subject: MATHEMATICS; University: University of Oklahoma; Term: Unknown 1989;

Typology: Quizzes

Pre 2010

Uploaded on 08/31/2009

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Math 3113, Quiz IV, Solution
1. Consider the initial value problem
tx000 + 2x00 + (t1)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))0sX (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)(s3s) + X(s)(s2s1) = 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 :
L1{1
s(s2+ 1)}=L1{1
s2+ 1}∗L1{1
s}= sin t1 =
t
Z
0
sin τ = 1 cos t.
(b) Using Partial Fractions :
1
s(s2+ 1) =A
s+Bs +C
s2+ 1 =1
ss
s2+ 1 L1{X(s)}= 1 cos t.
(c) Using Transform formula for integrals :
L1{1
s(s2+ 1)}=L1{
1
s2+1
s}=
t
Z
0
L1{1
s2+ 1} =
t
Z
0
sin τ = 1 cos t.
Hence, the solution is
x(t) = L1{X(s)}= 1 cos t.
1

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Math 3113, Quiz IV, Solution

  1. Consider the initial value problem

tx′′′^ + 2x′′^ + (t − 1)x′^ = 0 x(0) = x′(0) = x′′(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{tx′′′} = −(s^3 X(s))′^ = −(3s^2 X(s)+s^3 X′(s)), L{ 2 x′′} = 2s^2 X(s) and L{(t− 1)x′} = L{tx′} − L{x′} = −(sX(s))′^ − sX(s) = −(X(s) + sX′(s)) − sX(s). Hence applying Laplace Transform to the differential equation and collecting terms involving X(s) and X′(s) we get

X′(s)(−s^3 − s) + X(s)(−s^2 − s − 1) = 0 ⇒ X

′(s) X(s) =^ −

s^2 + s + 1 s^3 + 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.)

  1. Find the Laplace inverse of

X(s) = (^) s(s (^21) + 1)

There are several ways to solve this : (a) Using Convolution product Theorem :

L−^1 { (^) s(s (^21) + 1)} = L−^1 { (^) s (^2 1) + 1} ∗ L−^1 {^1 s } = sin t ∗ 1 =

∫^ t

0

sin τ dτ = 1 − cos t.

(b) Using Partial Fractions : 1 s(s^2 + 1)

= A

s

  • Bs^ +^ C s^2 + 1

=^1

s

− s s^2 + 1

⇒ L−^1 {X(s)} = 1 − cos t.

(c) Using Transform formula for integrals :

L−^1 { (^) s(s (^21) + 1)} = L−^1 {

1 s^2 + s }^ =

∫^ t

0

L−^1 { (^) s (^2 1) + 1}dτ =

∫^ t

0

sin τ dτ = 1 − cos t.

Hence, the solution is x(t) = L−^1 {X(s)} = 1 − cos t.