Solved: Section 5.2, page 202, use pencil and paper calculations and your table of Laplace transforms to pe

Section 5.2 Lecture Notes (Part 2)
The goal of this activity is to add a couple more important properties of the Laplace transform.

Theorem
Suppose f is piecewise continuous and of exponential order. If c is a constant, then

Proof: We’ll use our definition of the Laplace transform to write

where F is the Laplace transform of f. Now, many find the last step of the proof difficult to understand, so let’s
note that the Laplace transform of a function of t returns a function of s. That is,

where F is the Laplace transform of f. Now, let’s note what this means. Suppose we substitute 5 for s. Then we
would get

Now, note that if we substituted for s, we would get:

Let’s add this to our table.

1

Let’s try an example.

Example #1:
Find the Laplace transform of .

Solution: By property #4 in our table, we know that:

That is,

Therefore, by our new theorem (item #6 in our table), we can write:

Let’s check our answer with Matlab.

syms s t

2

y=exp(3*t)*cos(5*t);
Y=laplace(y)

Y =

Y=simplifyFraction(Y)

Y =

Same answer! Now, here is some visual evidence that it is true. Recall from algebra, that the graph of
is a shift of the graph of , c units to the right (provided ). Let’s use this fact to better understand what is
going on with our new property.

syms s t
g=cos(5*t);
G=laplace(g);
f=exp(3*t)*cos(5*t);
F=laplace(f);
fplot(G,[0,10])
hold on
fplot(F,[3,10])
grid on
xlabel(‘s-axis’)
title(‘Laplace transforms of g(t)=cos 5t and f(t)=exp(3*t)cos 5t’)
legend(‘F(s)’,’F(s-3)’,’Location’,’southeast’)
hold off

3

Nice! Clearly shows that if , then , because the graph of the second is translated 3
units to the right.

Theorem
Suppose f is a piecewise continuous function of exponential order. Let be the Laplace transform of f. Then,

Proof: Let’s start with the definition of the Laplace transform.

Now, let’s differentiate both sides with respect to s.

Note that we can bring the derivative with respect to s inside because we have an integral with respect to t.
Now, we can continue with:

4

Therefore, we’ve shown that:

Let’s try an example and see if this works.

Example #2:
Find the Laplace transform of .

Solution: Using item in our table, we know that if , then:

Therefore, by our new theorem,

so we’ll need to differentiate to find the answer.

Therefore,

Now, let’s check our answer with Matlab.

syms s t
y=t*exp(5*t);

5

Y=laplace(y)

Y =

Same answer! 🙂

Higher Powers of t
We’ve shown that

where F is the Laplace transform of f. Let’s find what is equal to.

Therefore, we’ve shown that

If we were to differentiate again, we would discover that:

If we were to differentiate again, we would discover that

where the notation means the “fourth derivative of F.” If we were to differentiate again, we would
discover that

If we were to differentiate again, we would discover that

6

There is a definite pattern developing here which can be finally expressed as

where means the “n th derivative of F.” Let’s add this to our table.

Let’s try an example.

Example #3:
Find the Laplace transform of .

Solution: Using item #5 in our table, we know that if , then

Therefore, by property #7 in our table,

Therefore, we’ll have to find the second derivative of . Let’s start with the first derivative.

7

We can check this answer with Matlab.

syms s t
F=4/(s^2+16);
diff(F,s)

ans =

Same answer, so we can find the second derivative.

A lot of work. Let’s check this derivative with Matlab.

diff(F,s,2)

ans =

simplifyFraction(diff(F,s,2))

ans =

Same answer. Therefore,

8

Now, we can quickly check our answer with Matlab’s laplace command.

y=t^2*sin(4*t);
Y=laplace(y)

Y =

Y=simplifyFraction(Y)

Y =

Same answer.

Let’s do another example. In our previous examples, we showed a number of detailed steps to help our readers,
but now we will start to take some shortcuts.

Example #4:
Write the following initial value problem in algebraic form, then solve for .

Solution: The Laplace transform of the left side of our equation is:

Note how we used the initial conditions and in our last step. Next, the right side. By property
#5 in our table,

Therefore, by property #6 in our table,

Therefore, the Laplace transform of our equation is:

9

Now we can solve for .

A lot of work, so errors are likely. Let’s check our answer with Matlab.

syms s t Y
f=exp(2*t)*sin(3*t);
F=laplace(f);
y0=1;
yp0=-1;
Y1=s*Y-y0;
Y2=s*Y1-yp0;
sol=solve(Y2-2*Y==F,Y);
sol=simplifyFraction(sol)

sol =

Perfect! 🙂

Example #5:
Write the following initial value problem in algebraic form, then solve for .

Solution: We’ll first find the Lapalce transform of the left side, using our initial conditions and .

Now we take the Laplace transform of the right side. First, by property #4 in our table, we have:

10

Therefore, by property #7 in our table,

so we need to differentiate .

We can check our differentiation with Matlab.

syms s
F=s/(s^2+9);
Fp=diff(F,s);
Fp=simplifyFraction(Fp)

Fp =

Same answer! Therefore, the Laplace transform of our right side is:

Now, we can set up an algebraic equation for our initial value problem with these results and solve for .

11

Wow! A lot of steps, so a mistake is likely. Let’s check our answer with Matlab. Recall that the initial value
problem is:

syms s t Y
f=t*cos(3*t);
F=laplace(f);
y0=1;
yp0=0;
Y1=s*Y-y0;
Y2=s*Y1-yp0;
sol=solve(Y2-2*Y1+Y==F,Y);
sol=simplifyFraction(sol)

sol =

Same answer! We’re done. 🙂

12

T 8. y(I): r:
” ro. y(il: e:,
* tr. )(/): cos3r

9. y(t) : 73
ll. y(tl: e-3′

/@ ,,,:,: sin5r

202 CHAPTER 5 ‘The Laplace Transform

Exercises 8-13 are designed to test the validity of Proposi-
tion 2. L In each exercise.
(i) cornpute 4(,y’)(s) for the given function, and
(ii) compute s 4(.v)(s) – )(0) for the given function. Com-

pare this result to that found in part (i) to verify that
t 4(,r,’)(.r): s4(.y)(s) *)(0).

39. y” I y’ + 2y : e-‘ cos2t, y(0) : l, y'(0) : -1
40. y” l2y’ + 5y : t2e-‘, y(0) : l, y'(0) : -2
41. y” *5y:3e-tcos4t,y(0) = -1,y'(0):)
42. Here is an interesting way to compute the Laplace trans-

form of cos ror.
(a) Using only Definition 1.1, show that f,{sint}(s) :

l/(s2 + l).
(b) Suppose that /(/) has Laplace transform F(s). Show

that

Llf (arlls) = j. (;)
Use this property and the result found in part (a) to
show that

^C{sina,lt}(s) : -3-.s”[email protected]
(c) If /(r) : sin cr.r/, then /'(r) : @ cos @t . Use Proposi-

tion 2.1 to show that

f{rocos<ot}(s): “: “s’+ a’
thus ensuring

J’
4,{cos ror}ts) : —,-;.s-+o’

-FIn
a manner similar to that proposed in Exercises 8-13, verify

-p the result of Proposition 2.4 for the functions defined in Exer-
– cises l4-17.E.-F r+. 1.(r) :11

N’ ro. y(r) : sin2r
15. .v(r) : s-2l

/fO )'(r) = rr + 3I + s
In Exercises 18-25, use Propositions 2.1,2.4, and2.7 to trans-
form the given initial value problem into an algebraic equation
involving,C(y). Solve the resulting equation for the Laplace
transform of v.,,A

” ).v’+ 3.v: 12. v(0) : -l19.y’-5y:””,.y(0):I
F ro. )’ +5y : t2 * 2r + 3.y(0) : 0
Zt. y’ – 4y : cos2r. y{0} : -2

22. y” * 1l : sin4/, y(0) :0, y'(0) : l
/@ y” *2y’ *2y : co, 2t, y(0): l, y'(0) :0

?* y” I y’ +2y : cos2t * sin3r, )(0) : -1, y'(0) : I
49 y” + 3y’r5y : t + e-t,y(o) : -1, y'(o) : o

In Exercises 26-29, we Proposition 2.12 to find the Laplace
transform of the given function.

2-6. y(t) : e-t sin3t 27. y(t) : e2′ cosZt
1Q,,r, : e-2t (2t + 3) 29. y(t) = e-‘(t2 + 3t + 4)

In Exercises 30-33, use Proposition 2.14 to find the Laplace
transform of the given function.
30. y(r): /sin3/ r,Qr(r): re-‘
32. y(t) : t2 cos2t 33. y(r) : 12r2t ‘

– In Exercise s 3441, use the propositions in Section 2 to trans-
-‘rt1 fo.rn the given initial value problem into an algebraic equation
( N involving LO). Solve the resulting equation for the Laplace
j L transform of y.
iIr. y’+2y: rsinI. y(o): t t/
S- 35. y’ – y : t2e-z’ ,y(0) : 0
– fO. y’ + ]l, : e-‘ sin 3r, l(O) : O Lr
l’ Sz. y’ – 2y – e2′ cos t, y(0) – -Z

=:

@tn”ro.
mafunctionis defined by

l-1a.1 : [- ,.’r’ ‘ ,lt. 1a > 0).
Jo

(a) Prove that f(1) : 1
(b) Prove that l(cvt 1) : af (cv). [n fact, if r is a positive

integer, show that f (r * 1) : n!.
(c) Show that

Llt’l(s’ : f’11 “‘{u1 r
‘

Indeed, ifr is a positive integer, use this result to shou
that,C{r”}(s) : nl/ s”+t.

44. Suppose that I is a continuous function for I > 0 and i.
of exponential order.
(a) If /(/) + F(s) is a transform pair, prove that

c[ [‘ rr,rar]t,r – F”’.lJu’ | .r
Hint: Let g(r) : Ll .f trlrlr. Then note that 8′(t) =
f (t) and use Proposition 2.1 to compute 4{g'(t)}(s t

(b) Use the technique suggested in part (a) to find

38. y” * 4y : t2 sin4t, y(0) :0, y'(0) : -1 “,1
I I’ lr,,rt+l)l’

_-l