IT Tallaght Maths Group: CC BY-NC-SA 3.0
Differential Equations Revisited
We have seen techniques for breaking a complex algebraic fraction down into
simpler parts, and finding the inverse Laplace Transform of these pieces. We will
now use these techniques to solve differential equations.
Example 1
The diagram shows a chicken in an oven. At time t the chicken temperature is y(t)
and the oven temperature is M(t).
Outside temp = M(t)
Newton’s Law of Cooling says that the
dy
rate of change of temperature
of a
dt
body at temperature y in an environment
of temperature M is proportional to M - y
(the difference in temperature.
i.e.
Temp = y(t)
dy
 k M  y 
dt
Where k is a constant related to the
thermal conductivity of the chicken.
Suppose that M (t )  3 sin( t ) i.e. the external temperature oscillates. Suppose the
chicken is such that k = 2 then at time t,
dy
 23 sin( t )  y 
dt

dy
 6 sin( t )  2 y
dt

dy
 2y  6 sin( t )
dt
… (1)
Suppose also that at time t = 0, y(0) = 50°. We want the temperature of the
chicken at time t.
(i) Show that Y (s ) 
6
s  2s 2  1

50
s  2
We have already seen that if Y is the Laplace Transform of y and Y  that of
then,
dy
dt
Y (s )  sY (s )  y (0)  sY  50
Using Laplace transforms to solve Differential Equations
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Laplace transforming the differential equation we have
Y   2Y  6.
1
s 1
2
 sY  50  2Y  6.
 s  2 Y 
Y 
6
s 1
6
2
1
s 1
2
 50
s  2s 2  1

50
s  2
Next need to find y(t).
Using partial fractions we get
6
1
6
s
12
1
1
Y .
 . 2
 . 2
 50.
s  2
5 s  2 5 s  1 5 s  1

 y (t ) 



6  2t 6
12
e
 cos(t )  sin( t )  50e 2t
5
5
5
For large t values, e 2t  0 so that for large t,
6
12
y (t )   cos(t )  sin( t )
5
5
i.e. the “steady state” temperature of the body also oscillates. If you plot y(t) and
the input temperature 6 sin(t ) you see that y(t) oscillates at the same frequency as
the input but has a smaller variation (amplitude) and “lags behind” it.
Example 2
In this problem you will solve the differential equation,
dx
 x  1  cos(t ) ,
x ( 0)  0
dt
(1) Show that if X is the Laplace Transform of x then,
X
1
s

ss  1 s  1 s 2  1


(2) Given that X can be rewritten as,
X (s ) 
 1 s  12 
1
1
1


 2
s s  1 2s  1
s2 1
Find the solution x(t) of the differential equation.
(3) What is the long term or “steady state” solution?
Using Laplace transforms to solve Differential Equations
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IT Tallaght Maths Group: CC BY-NC-SA 3.0
Solution
Using Laplace transforms to solve Differential Equations
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Example 3
Solve the differential equation
dy
 3y  1  0 ,
dt
Solution
y ( 0)  0
Using Laplace transforms to solve Differential Equations
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Example 4
Solve the differential equation
dx
 x   (t )  0 , x(0)  0
dt
Solution
Using Laplace transforms to solve Differential Equations
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Solution of 2nd Order Linear Differential Equations
A linear 2nd order differential equation with constant coefficients is of the form,
a
d 2y
dt
2
b
dy
 cy  f (t )
dt
where a, b and c are constants (i.e. independent of t). The function f(t) is called
the driving term. Such equations are used to model mechanical systems where y is
dy
d 2y
the position of a body,
its velocity and
its acceleration. The driving
dt
dt 2
term f(t) represents a force acting on the body.
Example 1
equilibrium level
y
A mass of 1kg hangs from a spring of spring constant k = 2. Ignoring air resistance,
the differential equation describing the displacement y at time t from the
equilibrium position is,
d 2y
dt 2
 2y  g
where g  10ms 2 is the acceleration due to gravity. Assuming the mass is
released from rest a distance of 0.1m below the equilibrium level, what is the
position of the mass at time t?
Solution
dy
0  velocity when t = 0, is 0. Released from
dt
rest initially at a distance 0.1m below equilibrium means y(0) = 0.1.
Released from rest means that
Let Y represent the Laplace Transform of y
Let Y  represent the Laplace Transform of
dy
dt
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d 2y
Let Y  represent the Laplace Transform of
Then transforming t
d 2y
dt 2
dt 2
 2y  10 gives
Y   2Y   10 s
… (1)
Now using Result 2 Y   sY  y (0)  Y   sY  0.1
and
dy
Y   sY  
(0 )
dt
Y   sY 
… (2)
 s sY  0.1 using (2) above for Y 
 s 2Y  0.1s
Substitute Y  and Y  into (1) gives,
s 2Y  0.1s  2Y  10
s
 s 2  2 Y  0.1s  10
s
 s 2  2 Y  10  0.1s
s
10
0.1s
Y 
 2
2
ss 2
s 2





Given that

10
ss 2
2


 

5
5s
then,
 2
s s 2
Y


5
5s
0.1s
 2
 2
s s 2
s 2

 

Looking at the table,
 
5
s
 5 and
 cos 2t
s
s2  2
 
 y (t )  5  4.9 cos 2t 
 
 y (t )  5  5 cos 2t  0.1cos 2t
i.e. y(t) oscillates either side of 5 with period
2.
Using Laplace transforms to solve Differential Equations
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Example 2
A mass on a spring is released in a viscous medium and the equation of motion is
d 2y
dt
2

4dy
 3y  10
dt
The initial position of the mass is y(0) = 2 and it is released from rest so that
dy
( 0)  0 .
dt
(1) If Y is the Laplace Transform of y show that, s 2Y  2s  4sY  8  3Y 
(2) Show that Y 

10
s

2 s 2  4s  5
s s  1s  3
(3) Use partial fractions to find Y and hence find y(t).
Using Laplace transforms to solve Differential Equations
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IT Tallaght Maths Group: CC BY-NC-SA 3.0
Example 3
In the first example, a friction term is added giving the equation,
d 2y
dt
2
2
dy
 2y  10 ,
dt
y (0)  0.1 and
dy
( 0)  0
dt
(1) Show that
s 2Y  2sY  2Y  0.1s  0.2  10
where Y is the Laplace Transform of y.
(2) Show that Y 
(3) Given that


10
s s 2s  2
2
10
s s  2s  2
2


0.1s  0.2
 s

s
2
 2s  2

5s  10
5
 2
s s  2s  2


show that, Y 
5 4.9s  9.8

s s 2  2s  2


(4) Complete the square for s 2  2s  2 and hence find y(t).
Using Laplace transforms to solve Differential Equations
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