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Diff Eq 7.2

 

<== Differential Equations
Number Question Answer
See the work
1. Consider the following initial value problem:
y'' -{4} y' - {21} y= \sin(4 t) \hspace{0.5in} y(0)=6, \; y'(0)=-1

Using Y for the Laplace transform of y(t), i.e., Y = \mathcal{L} \lbrace y(t) 
brace,
find the equation you get by taking the Laplace transform of the differential equation and solve for
Y(s) =

Y(s) =Y(s) =
The work

Point Cost: 3
2. Consider the following initial value problem:
y''-{2}y' - {24} y= \sin(8 t) \hspace{0.5in} y(0)=5, \; y'(0)=-3

Using Y for the Laplace transform of y(t), i.e., Y = \mathcal{L} \lbrace y(t) 
brace,
find the equation you get by taking the Laplace transform of the differential equation and solve for
Y(s) =

Y(s) =Y(s) =
The work

Point Cost: 3
3. Use the Laplace transform to solve the following initial value problem:
y''-{2}y' - {24} y= \sin(8 t) \hspace{0.5in} y(0)=5, \; y'(0)=-3

First, using Y for the Laplace transform of y(t), i.e., Y = \mathcal{L} \lbrace y(t) 
brace,
find the equation you get by taking the Laplace transform of the differential equation
=0

Now solve for Y(s) =
and write the above answer in its partial fraction decomposition, Y(s) = rac{A}{s+a}+rac{B}{s+b} where a < b
Y(s) = +

Now by inverting the transform, find y(t)= .
Differentialequation=
Y(s) =
Y(s) =+
y(t)=
The work

Point Cost: 3
4. Use the Laplace transform to solve the following initial value problem:
y''-{2}y' - {24} y= \sin(8 t) \hspace{0.5in} y(0)=5, \; y'(0)=-3

First, using Y for the Laplace transform of y(t), i.e., Y = \mathcal{L} \lbrace y(t) 
brace,
find the equation you get by taking the Laplace transform of the differential equation
=0

Now solve for Y(s) =
and write the above answer in its partial fraction decomposition, Y(s) = rac{A}{s+a}+rac{B}{s+b} where a < b
Y(s) = +

Now by inverting the transform, find y(t)= .
Differentialequation=
Y(s) =
Y(s) =+
y(t)=
The work

Point Cost: 3
5. Use the Laplace transform to solve the following initial value problem:
y'' +16 y = \cos (5 t) \hspace{0.5in} y(0)=0, \; y'(0)=0

First, using Y for the Laplace transform of y(t), i.e., Y = \mathcal{L} \lbrace y(t) 
brace,
find the equation you get by taking the Laplace transform of the differential equation and solving for Y:

Y(s) =

Find the partial fraction decomposition of Y(s) and its inverse Laplace transform to find the solution of the DE:

y(t)= .
Y(s) =
y(t)=
The work

Point Cost: 3
6. Use the Laplace transform to solve the following initial value problem:
x' = 12 x + 3 y, \; \; y'= -9 x + e^{3 t} \hspace{0.5in} x(0)=0, \; y(0)=0

Let X(s) = \mathcal{L} \lbrace  x(t) 
brace, and Y(s) = \mathcal{L} \lbrace  y(t) 
brace.
Find the expressions you obtain by taking the Laplace transform of both differential equations and solving for Y(s) and X(s):

X(s) =
Y(s) =
Find the partial fraction decomposition of X(s) and Y(s) and their inverse Laplace transforms to find the solution of the system of DEs:

X(s)
X(s) .
X(s) =
Y(s) =
X(s)
X(s)
The work
7. The inverse Laplace Transform of
X(s)
can be found in two ways.

METHOD 1:
Using Theorem 2 we have that f(t) = \int_0^t g(	au) d	au where g(t) = \mathcal{L}^{-1} \lbrace rac{4 s + 1}{s^2 + 25}  
brace.

a) Determine g(t) =

b) Integrate to find f(t) =

METHOD 2:
The Partial Fraction Decomposition of F(s) has the form:

rac{A}{s} + rac{ B s + C }{s^2 + 25}.
Determine A, B and C.
A =
B =
C =

Find f(t) using the Partial Fraction Decomposition.
f(t) =

g(t) =
f(t) =
A= 1/25
B= -1/25
C= 4
f(t) =

The work