# Review for the third midterm, Introduction to Diff. Eqs.,

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```					Review for the third midterm, Introduction to Diﬀ. Eqs., 3450:335-003, Dr. Montero, Spring 2009 Some useful facts regarding the Laplace transform. For a function f (t) deﬁned on [0, ∞), the Laplace transform L(f ) is the function of s deﬁned through the integral
∞

L(f )(s) =
0

e−st f (t) dt.

Here is an extremely short list of Laplace transforms of common functions: L(cos(ωt))(s) = L(cosh(ωt))(s) = s ω , L(sin(ωt))(s) = 2 , s2 + ω 2 s + ω2 s2

s ω , L(sinh(ωt))(s) = 2 , 2 −ω s − ω2 n! 1 . L(tn )(s) = n+1 , L(eat )(s) = s s−a The following formulas give the transform of functions related to f (t) in terms of the transform of f (t): L(eat f (t))(s) = L(f )(s − a), L(tf (t))(s) = − L(f (t))(s) = sL(f )(s) − f (0), L f (n) (t) (s) = sn L(f )(s) − sn−1 f (0) − sn−2 f (0) − ... − sf (n−2) (0) − f (n−1) (0), L (tn f (t)) (s) = (−1)n dn L(f )(s). dsn d L(f )(s), ds

Example: Find the solution of the following initial value problem: y − y − y + y = 0, y(0) = y (0) = 0, y (0) = 1.

Solution: We apply L to the equation above. Writing Y (s) = L(y), we obtain s3 Y − 1 − s2 Y − sY + Y = 0. This means that Y (s) = s3 − s2 1 1 1 1 = 2 = 2 = . −s+1 s (s − 1) − (s − 1) (s − 1)(s − 1) (s − 1)2 (s + 1)

To determine y(t) from here ﬁrst need to split this into partial fractions. Here is what I get: 1 1 1 1 1 1 1 = − − . 2 (s + 1) (s − 1) 4 s + 1 4 s − 1 2 (s − 1)2 The inverse transform of the ﬁrst two terms is easy. For the last we notice that 1 d 1 =− = L(tet )(s). 2 (s − 1) ds s − 1

We conclude that

1 1 1 y(t) = e−t − et − tet . 4 4 2

Example: Find the anti-transform of F (s) = s2 2s + 5 . + 6s + 34

Solution: We ﬁrst write F (s) in a way amenable to our computations: 2s + 5 2s + 5 2(s + 3) − 1 s+3 1 5 = = =2 − . 2 + 25 2 + 25 2 + 25 + 6s + 34 (s + 3) (s + 3) (s + 3) 5 (s + 3)2 + 25

s2

From here we obtain 1 L−1 (F (s)) = 2e−3t cos(5t) − e−3t sin(5t). 5

Convolution. For two functions f , g deﬁned on the interval [0, +∞), the convolution of f and g, denoted by f ∗ g, is the following function:
t

(f ∗ g)(t) =
0

f (x)g(t − x) dx.

A key property of the convolution is the following: If F (s) = L(f )(s) and G(s) = L(g)(s), then L(f ∗ g)(s) = F (s)G(s), that is, the Laplace transforms converts a convolution into a product. Example: Find the anti-transform of H(s) = 4s . s4 + 5s2 + 4

Solution: We notice that H(s) = (s2 4s 2 s =2 2 × 2 = 2L (sin(2t) ∗ cos(t)) . 2 + 1) + 4)(s s +4 s +1

From here we conclude that h(t) = L
−1

4s 2 + 4)(s2 + 1) (s

t

= 2 sin(2t) ∗ cos(t) = 2
0

sin(2x) cos(t − x) dx.

Now we need to compute the last integral. In this case the following identities are helpful: cos(α ± β) = cos(α) cos(β) and sin(2β) = 2 sin(β) cos(β). From here we get
t t

sin(α) sin(β),

sin(2x) cos(t − x) dx =
0 0

sin(2x) (cos(t) cos(x) + sin(t) sin(x)) dx
t t

= cos(t)
0

sin(2x) cos(x) dx + sin(t)
0 t

sin(2x) sin(x) dx
t

= 2 cos(t)
0

cos (x) sin(x) dx + 2 sin(t)
0

2

sin2 (x) cos(x) dx.

The last two integrals are easy to compute using a substitution. We obtain h(t) = 4 4 cos(t)(1 − cos3 (t)) + sin(t) sin3 (t). 3 3

Unit Step. The unit step function, usually denoted by u(t), is deﬁned as follows: u(t) = 0 t<0 1 t≥0

Note that, for a > 0, the function u(t − a), which is often denoted by ua (t), corresponds to ua (t) = u(t − a) = 0 t<a 1 t≥a

The unit step gives us the following formula, valid for a > 0: L(f (t − a)u(t − a))(s) = e−as L(f )(s). In particular, we obtain the following: L(u(t − a))(s) = e−as . s

Example: Find the solution of the following initial value problem: y − 6y + 8y = h(t), where h(t) = 0 t<8 (t − 8) t ≥ 8 y(0) = y (0) = 0,

Solution: We ﬁrst need to ﬁnd a usable expression for the function on the right hand side. In this case the function is g(t) = t, translated by 8 units to the right, that is g(t − 8), and it is also 0 for t < 8. In other words, h(t) = u(t − 8)g(t − 8), where g(t) = t. hence e−8s . s2 Writing Y (s) = L(y)(s), we transform the equation to obtain L(h) = (s2 − 6s + 8)Y = so e−8s , s2

e−8s . Y (s) = 2 2 s (s − 6s + 8)

To obtain the inverse transform of this expression we notice ﬁrst that s2 (s2 1 1 = 2 − 6s + 8) s (s − 4)(s − 2) 1 1 1 − = 2 2s s − 4 s − 2 1 1 1 1 = 2× − 2× . 2s s − 4 2s s−2 1 =L − 6s + 8) 1 1 t ∗ e4t − t ∗ e2t . 2 2

This says that s2 (s2

Here we need to compute the convolution
t t

t ∗ eαt =
0

xeα(t−x) dx = eαt
0

xe−αx dx = −

t 1 + 2 (eαt − 1), α α

where the I integrated the last integral by parts. So far we have s2 (s2 This ﬁnally says that e−8s (t − 8) 1 (t − 8) 1 2(t−8) = L u(t − 8) − + (e4(t−8) − 1) + − (e − 1) s2 (s2 − 6s + 8) 8 32 4 8 which is the same as saying L−1 e−8s s2 (s2 − 6s + 8) = u(t − 8) − 1 (t − 8) 1 2(t−8) (t − 8) + (e4(t−8) − 1) + − (e − 1) . 8 32 4 8 , t 1 t 1 1 = L − + (e4t − 1) + − (e2t − 1) . − 6s + 8) 8 32 4 8

Derivative of a Transform. When the relation between two functions is of the form g(t) = tk f (t) for some integer k ≥ 1, there is a very simple way to obtain the Laplace transform of g from that of f : L (tn f (t)) (s) = (−1)n dn L(f )(s). dsn

Example: Find the inverse transform of F (s) = ln(s − 2) − ln(s − 4). Solution: Let us call f (t) the function with L(f ) = F (s). Then L(tf )(s) = − This obviously says that tf (t) = e2t − e4t , or f (t) = e2t − e4t . t d 1 1 F (s) = − = L(e2t − e4t )(s). ds s−2 s−4

Transform of a Periodic function. A function f (t) is called periodic of period T > 0 if f (t + T ) = f (t) for all t. The following formula is valid for f (t) periodic of period T : 1 L(f )(s) = 1 − e−T s
T

e−st f (t) dt.
0

Example: Compute the Laplace transform of the function in Figure 1, that is, of the function that is periodic, with period T = 2, and for which h(t) = sin πt 4
1

when 0 ≤ t < 2.
2 4 6

Solution: We know that h(t) is periodic with period T = 2, so we have L(h)(s) = 1 1 − e−2s
2

Figure 1: h(t).

e−st h(t) dt =
0

1 1 − e−2s

2

e−st sin
0

πt 4

dt.

To compute this integral we can integrate by parts twice, or look in a table of integrals: eαt sin(ωt) dt = eαt (α sin(ωt) − ω cos(ωt)) . α2 + ω 2

We will use this formula for α = −s and ω = π , to obtain 4
2

e−st sin
0

πt 4

dt = = =

e−st s2 + 1 s2 + (16s2
π 2 4 π 2 4

−s sin

πt 4 π 4

−

π πy cos 4 4

2 0

−se−2s +

4 −4se−2s + π . + π2)

All this together tells us that L(h)(s) = 1 4 −4se−2s + π . −2s (16s2 + π 2 ) 1−e

Dirac Delta. The Dirac Delta, or unit impulse function, δ(t − t0 ) for a ﬁxed number t0 > 0, is an object usually associated with the following properties: (i) δ(t − t0 ) = 0 for all t = t0 (ii)
∞ −∞

δ(t − t0 ) dt = 1.

The Laplace transform of δ(t − t0 ) is L(δ(t − t0 ))(s) = e−t0 s . Example: Solve the following initial value problem: y + 4y + 13y = δ(t − 2) − δ(t − 4), y(0) = y (0) = 0.

Solution: Write Y (s) = L(y)(s). Applying L to the equation we obtain (s2 + 4s + 13)Y = e−2s − e−4s . This says that Y (s) = e−4s 1 e−2s − = (s + 2)2 + 9 (s + 2)2 + 9 3 L e−2t sin(3t) (s) = In light of the unit step formula, we obtain y(t) = 1 −2(t−2) e sin(3(t − 2)) − e−2(t−4) sin(3(t − 4)) . 3 3e−2s 3e−4s − (s + 2)2 + 9 (s + 2)2 + 9 3 . (s + 2)2 + 9 .

We next notice that

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