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Material Type: Quiz; Class: Introduction to Ordinary Differential Equations; Subject: MATHEMATICS; University: University of Oklahoma; Term: Unknown 1989;
Typology: Quizzes
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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.)
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)
s
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.