Solutions to homework #10
Ÿ Problem 7.2.13
Ÿ Hand calculation and table lookup
Compute LA6 e-3 t - t2 + 2 t - 8E. The first step is to use the linearity of the Laplace transform to distribute the transform over the
sum and pull the constant factors outside the transforms. That gives
LA6 e-3 t - t2 + 2 t - 8E = 6 LAe-3 t E - LAt2 E + 2 L@tD - 8 L@1D.
Using the book’s table, we find
1
LAe-3 t E = s+3
€€€€€€€€€ ,
2
LAt2 E = €€€€€3 ,
s
1
L@tD = €€€€€2 ,
s
(entry 14 with a = -3)
(entry 15 with n = 2)
(entry 15 with n = 1)
and
1
L@1D = €€€s
(entry 14 with a = 0).
Therefore
6
2
2
s
s
8
LA6 e-3 t - t2 + 2 t - 8E = s+3
€€€€€€€€€ - €€€€€3 + €€€€€2 - €€€s .
Ÿ Calculation with Mathematica
We can also compute the Laplace transform automatically using Mathematica,
In[2]:=
Out[2]=
LaplaceTransform@6 Exp@-3 tD - t ^ 2 + 2 t - 8, t, sD
2
2
6
8
€€€€€€€€ - €€€€€€€€ + €€€€€€€€€€€€€€€€€ - €€€€€
s2 s3 s + 3 s
which is identical to the result we got using table lookup.
2 hwk10Soln.nb
Ÿ Problem 7.2.15
Ÿ Hand calculation and table lookup
Compute LAt3 - t et + e4 t cos tE. The first step is to use the linearity of the Laplace transform to distribute the transform over the
sum and pull the constant factors outside the transforms. That gives
LAt3 - t et + e4 t cos tE = LAt3 E - L@t et D + LAe4 t cos tE.
Using the book’s table, we find
6
LAt3 E = €€€€€4 ,
(entry 15 with n = 3)
s
1
L@t et D = €€€€€€€€€€€€€€€2 ,
(entry 15 with n = 1 and a = 1)
Hs-1L
and
s-4
LAe4 t cos tE = €€€€€€€€€€€€€€€€€€€€€
€
2
(entry 26 with a = 4 and b = 1).
Hs-4L +1
Therefore
6
1
s-4
s
Hs-1L
Hs-4L +1
LAt3 - t et + e4 t cos tE = €€€€€4 - €€€€€€€€€€€€€€€2 + €€€€€€€€€€€€€€€€€€€€€
€.
2
Ÿ Calculation with Mathematica
We can also compute the Laplace transform automatically using Mathematica,
In[3]:=
Out[3]=
LaplaceTransform@t ^ 3 - t Exp@tD + Exp@4 tD Cos@tD, t, sD
s-4
1
6
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ - €€€€€€€€€€€€€€€€€€€€€€€€€€ + €€€€€€€€
2
2
s4
Hs - 4L + 1 Hs - 1L
which is identical to the result we got using table lookup.
Ÿ Problem 7.4.3
Ÿ Hand calculation and table lookup
Find the inverse Laplace transform of
s+1
FHs_L = €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
s2 + 2 s + 10
s+1
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
2
s + 2 s + 10
This expression isn’t in the table, so we’ll need to do some algebra to get it into one of the forms appearing in the table of Laplace
transforms. If the denominator can be factored, then we’d do a partial fraction expansion to split it into simpler terms of the form
1
€€€€€€€€€
s+a
which we’d recognize (from entry #14 in the book’s table) as the Laplace transform of an exponential. However, if the
denominator is irreducible over the reals (i.e., can’t be factored into reals) then we’d complete the square to get it into the forms
appearing in entries #25 and #26 in the table.
So the first question to ask is whether the denominator can be factored in real numbers. To be factored in reals, the discriminant
b^2 - 4 a c must be positive. Computing the discriminant with a=1, b=2, and c=10 gives the value -36, so the denominator
doesn’t factor in real numbers. Therefore we proceed to complete the square.
2
Recall that to complete the square we write s2 + b s + c in the form Hs + AL + B2 . Expanding the square gives s2 + 2 A s + A2 + B2 .
We then equate coefficients of each power of s (which we can do because the powers of s are linearly independent functions) to
find the equations
transforms. If the denominator can be factored, then we’d do a partial fraction expansion to split it into simpler terms of the form
1
€€€€€€€€€
s+a
which we’d recognize (from entry #14 in the book’s table) as the Laplace transform of an exponential. However, if the
hwk10Soln.nb
denominator is irreducible over the reals (i.e., can’t be factored into reals) then we’d complete the square to get it into
the forms
appearing in entries #25 and #26 in the table.
3
So the first question to ask is whether the denominator can be factored in real numbers. To be factored in reals, the discriminant
b^2 - 4 a c must be positive. Computing the discriminant with a=1, b=2, and c=10 gives the value -36, so the denominator
doesn’t factor in real numbers. Therefore we proceed to complete the square.
2
Recall that to complete the square we write s2 + b s + c in the form Hs + AL + B2 . Expanding the square gives s2 + 2 A s + A2 + B2 .
We then equate coefficients of each power of s (which we can do because the powers of s are linearly independent functions) to
find the equations
b=2A
and
c = A2 + B2 .
2
With b = 2 and c = 10 we find A = 1 and B = 3. To check a completion of squares, plug A and B into Hs + AL + B2 and expand it
out. You might want to try it by hand; it’s also easy to do in Mathematica:
2
ExpandBHs + 1L + 32 F
s2 + 2 s + 10
which is the original polynomial. Therefore we’ve completed the squares correctly, and we can rewrite FHsL as
1+s
€€€€€€€€€€€€€€€€€€€€€€€€
€.
2
H1+sL +32
This is entry #26 in the table on the book’s back cover, so we can recognize it as the Laplace transform of f HtL = e-t cos 3 t.
Ÿ Calculation in Mathematica
I should also point out that you can do the whole works in Mathematica using the InverseLaplaceTransform function:
ILT@t_D = InverseLaplaceTransform@F@sD, s, tD
1
€€€€€ ãH-1-3 äL t I1 + ã6 ä t M
2
f@t_D = FullSimplify@ILT@tDD
ã-t cosH3 tL
which is identical to the result found by algebraic manipulation and table lookup.
Finally, we can check by computing the Laplace transform of the answer
LaplaceTransform@Exp@-tD Cos@3 tD, t, sD
1+s
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
2
9 + H1 + sL
Together@ExpandAll@%DD
1+s
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
10 + 2 s + s2
which is the FHsL with which we started. Therefore our calculation checks.
4 hwk10Soln.nb
Ÿ Problem 7.4.5
Ÿ Hand calculation and table lookup
Find the inverse Laplace transform of
1
FHs_L = €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
s2 + 4 s + 8
1
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
s2 + 4 s + 8
The denominator doesn’t factor in the reals (which you can check) so we complete the squares. Using the procedure for complet2
ing the square shown in the previous problem, the denominator can be rewritten as Hs + 2L + 4 (which you should also check).
Looking up
1
FHs_L = €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
2
Hs + 2L + 4
1
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
2
Hs + 2L + 4
1
1
This is close to entry #25 in the table; it differs by a factor of 2€€€ . Thus the inverse Laplace transform of FHsL is f HtL = 2€€€ e-2 t sin 2 t.
Check by computing the Laplace transform:
LaplaceTransform@1  2 Exp@-2 tD Sin@2 tD, t, sD
1
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
2
Hs + 2L + 4
ExpandAll@%D
1
€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
s2 + 4 s + 8
which recovers the starting point. Therefore the calculation checks
Ÿ Calculation with Mathematica
Alternatively, we can use the InverseLaplaceTransform function to do the problem in one shot:
ILT@t_D = InverseLaplaceTransform@1  Hs ^ 2 + 4 s + 8L, s, tD
1
- €€€€€ ä ãH-2-2 äL t I-1 + ã4 ä t M
4
FullSimplify@ILT@tDD
1
€€€€€ sinH2 tL HcoshH2 tL - sinhH2 tLL
2
This looks different, but using the definition of the hyperbolic functions you can show that it is the same as the result we got using
table lookup.
Ÿ Problem 7.5.15
hwk10Soln.nb 5
Problem 7.5.15
Ÿ Table lookup & hand calculation
Solve for YHsL = L@yD given y’’ - 3 y’ + 2 y = cos t with initial conditions yH0L = 0 and y’ H0L = -1. Consulting our friendly table,
we find
L@y’’D = s2 Y - s yH0L - y’ H0L
(entry 4)
L@y’D = s Y - yH0L
(entry 3)
so that the Laplace transform of the LHS is
L@y’’ - 3 y’ + 2 yD = Is2 - 3 s + 2M Y + 1.
We can also look up the Laplace transform of the RHS,
s
L@cos tD = €€€€€€€€€€€2€ .
1+s
Therefore
s
Is2 - 3 s + 2M Y + 1 = €€€€€€€€€€€2€
1+s
from which we can solve for YHsL,
-1
s
s -3 s+2
Is -3 s+2M Is +1M
YHsL = €€€€€€€€€€€€€€€€€€€€€
€ + €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
€.
2
2
2
Adding the two terms gives
s-1-s2
YHsL = €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€
€.
2
2
Is -3 s+2M Is +1M
Ÿ Calculation with Mathematica
Here’s how to do it semi-automatically. First compute the Laplace transform of the LHS
In[5]:=
Out[5]=
LHS = LaplaceTransform@y ’’@tD - 3 y ’@tD + 2 y@tD, t, sD
HLt @yHtLD HsLL s2 - yH0L s + 2 HLt @yHtLD HsLL - 3 Hs HLt @yHtLD HsLL - yH0LL - y¢ H0L
Because Mathematica doesn’t yet know the initial conditions we have to use transformation rules to replace yH0L and y’ H0L with
the initial values. Furthermore, the Laplace transform of the unspecified function yHtL can’t be computed; calling the LaplaceTransform function on it simply returns a "placeholder" expression:
In[9]:=
Out[9]=
LaplaceTransform@y@tD, t, sD
Lt @yHtLD HsL
Therefore I’ll also use a transformation rule to replace LaplaceTransform[y[t], t, s] with YHsL. In Mathematica, this is done as
follows:
In[6]:=
LHS = LHS . 8y ’@0D ® -1, y@0D ® 0, LaplaceTransform@y@tD, t, sD ® Y@sD<
resulting in
Out[6]=
YHsL s2 - 3 YHsL s + 2 YHsL + 1
Now I take the Laplace transform of the RHS
In[7]:=
Out[7]=
RHS = LaplaceTransform@Cos@tD, t, sD
s
€€€€€€€€€€€€€€€€€€€€€
2
s +1
Having computed the Laplace transforms of both sides, we can set up an equation for YHsL,
Ÿ
6 hwk10Soln.nb
Having computed the Laplace transforms of both sides, we can set up an equation for YHsL,
In[10]:=
Out[10]=
eqn = LHS Š RHS
s
YHsL s2 - 3 YHsL s + 2 YHsL + 1 ‡ €€€€€€€€€€€€€€€€€€€€€
s2 + 1
which we can solve for YHsL
In[8]:=
Out[8]=
Solve@LHS Š RHS, Y@sDD
-s2 + s - 1
::YHsL ® €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ >>
Hs2 + 1L Hs2 - 3 s + 2L
This is identical to the result computed by hand.