Answers to the first set of review questions

solution to pg 232 # 2

```        {0,        0< t <2     }
f(t)=  {1,        2< t < 4    }
{0,        t> 4         }
```
f(t)= 0 + U(t-2) -U(t-4)
L(f(t))= (e^(-2s))*(1/s) - (e^(-4s))*(1/s)
F(s)= (e^(-2s) - e^( -4s))/s

7) Laplace Transform of e^(-7t) = 1/(s+7) 8) Laplace Transform(t*e^(-7t)) = [1/(s+7)^2].

Problem 9: L(sin2t) = 2/(s^2+4) using the formula L(sinkt) = k/(s^2 + k^2)

Problem 10: L(exp(-3t)sin(2t)) = ?

This fits the form exp(at)f(t) exp(at) = exp(-3t) => a = -3 f(t) = sin(2t)

L(exp(at)f(t)) = F(s-a)

L(sin(2t)) = 2/(s^2+4) (as shown in prob. 9)

therefore, L(exp(-3t)sin(2t)) = 2/(s^2+4) evaluated at s-a = s+3 = 2/[(s+3)^2 +4]

p.232, 13 & 14

13) Taking inverse laplace of 20/s^6, we see that it fits the form n!/s^(n+1). The n! is not in the numerator, so we have to multiply and divide by 5!. The invlaplace( 5!/s^6)=t^n=t^5. Multiplying by the constant 20 and dividing by 5! gives t^5/6.

14) We need to get this funtion into the form 1/(s-a). Therefore we multiply numerator and denominator by 1/3. This yields (1/3)/(s-(1/3)) and the inverse laplace is exp(at) where a=(1/3) in this case. Therefore we have (1/3)exp(t/3).

15) from the inverse laplace of 1/(s-5)^3, we can see that it involves the shifting in the s space, which means it fits the form of exp(at)*f(t).
Since a=5, we have exp(5*t)*f(t). The inverse of 1/s^3 fits the form of n!/s^(n+1). Since n=2, we will have t^2 but we still need to divide by 2! to compensate the extra factor. Thus, the result is 1/2*exp(5*t)*t^2.

16) the inverse laplace 1/(s^2-5) can be written as 1/(s^2-sqrt(5)^2) which fits the form of k/(s^2-k^2). We can see k=sqrt(5).
So the function is sinh(sqrt(5)*t), but we still need to divide by sqrt(5) to compensate [for] the extra factor. Therefore, the result is 1/sqrt(5)*sinh(sqrt(5)*t).

L^(-1){s/(s^2-10s+29)} =
L^(-1){ (s-5/((s-5)^2+4)) + (5/((s-5)^2+4))} =
e^(5t)*cos(2t) + (5/2)*e^(5t)*sin(2t)

L^-1 {(e^(-5s)) / s^2} = L^-1 {(1/s^2) * e^(-5s)} ==>

==>   L^-1 {e^(-as) * F(s)} = f(t-a) * U(t-a)

Here a=5

So:

L^-1 {(e^(-5s)) / s^2} = (t-5) * U(t-5)

-Joe Salvino-

19. Inverse Laplace Transform of [(s+pi)/(s^2+pi^2)]*(e^-s)
= L^-1 of [(s*e^-2)/(s^2+pi^2) + (pi*e^-2)/(s^2+pi^2)]

The function is of the form [e^-(a*s)]*F(s)
Therefore the inverse Laplace transform has the form f(t-a)*U(t-a)
In this case f(t) = cos(pi(t-1)) * U(t-1) + sin(pi(t-1)) * U(t-1).

```p233 #25-28
25.  y = U(t-to)f(t)
26.  y = f(t)-U(t-to)f(t)
27.  y = f(t-to)U(t-to)
28.  y = f(t) - f(t)U(t-to) + f(t)U(t-t1)
```

f(t)= sin t U(t-pi)-sin t U(t-3pi)
L{f(t)} =s (e^-[pi]s - e^-3[pi]s) / s^2 +1
L{ e^t f(t)} = (s-1) [ e^-pi(s-1) - e^3pi(s-1) ]/(s-1)^2 +1

Express f in term of unit step functions and the find L{f(t)} and L{e^t*f(t)}.

Solution:

According to the graph on figure 4.61, we have the following:
f(t)= 2 for 0<t<2 and f(t)= t for t>=2
Since u(t-2)= 0 for 0<t<2 and u(t-2)=1 for t>1, we have:
f(t)= 2- 2*u(t-2) + t*u(t-2). Let's find F(s)=L{f(t)}
Let's put g(t)=2 and h(t)=t. So, f(t)= g(t) - g(t)*u(t-2) + h(t)*u(t-2)
=> L {f(t)}=L{g(t)} - L{g(t)*u(t-2)} + L{h(t)*u(t-2)}
=> F(s) = G(s) - e^(-2s)*L{g(t+2)}+e^(-2s)L{h(t+2)}
g(t)=2 => G(s)= 2/s
g(t+2)=2 => L{g(t+2)}= 2/s
h(t+2) = t + 2 => L{h(t+2)} = 1/(s^2) + 2/s
so, F(s) = 2/s - e^(-2s)*(2/s) + e^(-2s)*( 1/(s^2) + 2/s)
=> F(s) = 2/s + e^(-2s) * (1/s^2)
Lets find L{e^t*f(t)}
L{e^t*f(t)}= F(s-1) = 2/(s-1) + e^(-2(s-1)) * (1/(s-1)^2)

f(t) = t - t U (t - 1)+ (-t +2) U (t - 1) - (-t +2) U(t -2)
F(S)= 1/s^2 - e^-s/(s+1)^2 + e^-s (3/s - 1/s^2) - e^-2s (4/s -1/s^2)
Laplace transfor of e^t*f(t)
1/(s-1)^2 - e^(-s-1)/s^2 + e^(-s-1) ((3/(s-1) - 1/(s-1)^2)) - e^(-2s-1)((4/(s-1) - 1/(s-1)^2))

Review Problem #35

```y"+6y'+5y = t -tU(t-2)
s^2Y -sY(0) -y'(0) +6sY -6y(0) +5y= 1/s^2 - exp(-2s)(1/s^2  +2/s)```
Rearrange
Y= 1/(s^2(s+5)(s+1)) +1/(s+5)(s+1) -(exp(-2s)+2s*exp(-2s))/(s^2(s+5)(s+1))

Partial Fractions

```Ex: for the 1/(s^2(s+5)(s+1)) term
A/s +B/s^2 +C/(s+5) +D/(s+1)
s(s+5)A +(s+5)(s+1)B +s^2(s+1)C +s^2(s+5)D=1
if s=0; B=1/5
if s= -5; C= -1/100
if s= -1; D= 1/4
if s=1; A= -6/25
```
Take the inverse laplace
```L^-1(1/s)= constant
L^-1(1/s^2)=t
L^-1(s+1)=exp(-t)
L^-1(s+5)=exp(-5t)```
Combine like terms
```-6/25 +t/5  +(3/2)exp(-t) -(13/50)exp(-5t) -(4/25)U(t-2)
-(1/5)(t-2)U(t-2) +(1/4)exp(-(t-2))U(t-2) -(1/5)exp(-5(t-2))U(t-2)```
However, I got:
```-6/25 +t/5  -(3/2)exp(-t) -(437/50)exp(-5t) -(4/25)U(t-2)
-(1/5)(t-2)U(t-2) +(1/4)exp(-(t-2))U(t-2) -(1/5)exp(-5(t-2))U(t-2)```
Note from the instructor
Maple got
`1/5*t-6/25-13/50*exp(-5*t)+1/2*exp(-t)-Heaviside(t-2)*t`
Sigh.

Problem 37:

First transforming the right side derivative we get: (F(s)*s)-1
>From the definition of the Volterra Equation we see that that right side is equal to: L{cos(t)} + [F(s) * L{cos(t)}]
Solving for F(s) we get: F(s) = (s^2 + s +1)/(s^3)
Using partial fractions we get A=1,B=1,C=1 Then, by inverse Laplace: L^-1{(1/s) + (1/s^2) + (1/s^3)}
Answer: 1 + t + .5t^2

write (1,5,3,4) as a linear combination of (2,1,0,1) (3,-1,1,0) and (0,1,2,3) in four space

first create the equation: A(2,1,0,1) + B(3,-1,1,0) + C(0,1,2,3) = (1,5,3,4)

this corresponds to the linear equations:

```2A + 3B + 0C = 1
A - B + C = 5
0A + B + 2C =3
A + 0B + 3C = 4```
these equations form an augmented matrix and when we try to reduce to RREF , by row operations, we will find the linear combination:
```2  3  0  1            2  3  0  1            2  3  0  1
1 -1  1  5            1 -1  1  5            1 -1  1  5
0  1  2  3            0  1  2  3            0  1  2  3
1  0  3  4            0  1  2 -1            0  0  0 -4
-row2+row4            -row3+row4```
after these two row operations are done we find their is no solution, shown in the fourth row. So there is no linear combination of these vectors which equals the vector (1,5,3,4).

Maintained by greenfie@math.rutgers.edu and last modified 10/6/2004.