Solving Differential Equations
Where a Dirac Delta Function exists at the Origin
We’ll limit this discussion to second order differential equations only and, for the moment, where the
input experiences a single derivative.
d2
d
D y  t   a D y  t   b y  t   c0 2 x  t   c1 x  t   c2 x  t 
dt
dt
y (0)  ?
(1.1)
2
where D 
d
dt
y '(0)  ?
The solution to this differential equation will be broken into two parts, the zero-state, yzs  t  , and the
zero-input solutions, yzi  t  . The final solution will then be
y  t   yzs  t   yzi t 
(1.2)
Zero-Input Solution
The zero-input solution is the same as covered previously and will not be repeated. If you need
assistance, please refer back to the earlier paper.
Zero-State Solution
To find the zero-state solution we’ll first have to find the impulse response of the system, h  t  , and
then convolve it with the input x  t  . To accomplish this we’ll have to progress through several steps:
1. Rewrite equation (1.1)
 D 2  a D  b  y  t   c0 D 2  c1D  c2  x  t 
Q D 
P D 
2. Find h0  t  in the equation below that is made up of the characteristic modes of the system
(e.g., the General Solution forms) where all the initial condition at 0  , just before the
application of the delta function at the origin, are zero:
Q( D) h0  t   D 2 h0  t   a D h0  t   b h0  t     t 
h0 (0 )  0
(1.3)
D h0 (0 )  0
Since the delta function at the origin changes the initial conditions, we cannot use the initial
conditions provided since they are before the origin. What is needed are the initial conditions
just after the origin, e.g. where t  0  . To obtain them we’ll use a heuristic argument.
To get the delta function at the origin it must have come from the highest order differential on
the left side of the equation. For the above equation, it is
order differential, such as
d2
h0  t  . To come from a lower
dt 2
d
d2
h0  t  , it would mean that 2 h0  t  would produce a doublet
dt
dt
which doesn’t exist on the right side of the equation. Knowing that the delta function came
from
d
d2
h t means that h0  t  must produce a step function. This give us the following
2 0 
dt
dt
equation just after the origin:
D 2 h0  t   a D h0  t   b h0  t   0
h0 (0 )  0
(1.4)
D h0 (0 )  1
Using the process used before to find the General Solution’s form, e.g., the characteristic modes
of the system, you can use the two initial conditions to find h0  t  .
3. The impulse response of the original system, h  t  , can be found using the following formula:
h  t   c0   t   P  D h0  t 
(1.5)
A caveat on the term c0   t  . This term appears only when P[D] has the same order as Q[D].
For example, if Q[D] is a second order differential equation, P[D] must also be second order or
c0   t   0.
4. The zero-state solution can now be found via convolution
yzs  t   x  t   h  t 

(1.6)

 x   h  t    d

Example 1
To find the total response of this system you must use the convolution between the input x  t 
and the impulse response h  t  since there is a derivative of the input
d2
d
d
d
y
t

3
y
t

2
y
t

x
t

2
x
t
where
y
0

6
and
y 0  5












dt 2
dt
dt
dt
x  t   e 3t u  t 
Step 1:
d2
d
Q  D  2  3  2
dt
dt
d
P  D   2
dt
Step 2:
 d2

d

3

2
 dt 2
 h0  t   0 t  0
dt


h0 (0 )  0
d
h0 (0 )  1
dt
This give the characteristic equation as
 2  3  2  0
or    1   2   0
The form of h0  t  is
h0  t   K1et  K2e2t
Using the initial conditions to solve for K1 and K 2 gives
h0  t   0  K1  K 2

 K1  1, K 2  1
h0  t   1   K1  2 K 2 
Therefore
h0  t   et  e2t
t 0
. Since the order of the differential equation is not the same (2nd order on the left and 1st order
on the right), h  t  will not contain an impulse (e.g., c0  0 ). The impulse response is then
h  t   P  D  h0  t 
d

   2   e t  e 2t  t  0
 dt

t
e
t 0
A couple of things to note. First, to find the impulse response we have not had to use either x  t  nor
the initial conditions given for the system, y  0  and y  0  . You will, however, need these to find the
total solution, y  t   yzs  t   yzi  t  !
First, to find the zero-state solution, you have to convolve x  t  with the just derived impulse response,
h t  .
y zs  t   x  t   h  t 

  x   h  t    d
0


 e
3

t
u    e t  u  t     d


  e 3 e t   d
0

1  t 1 3t
e  e
t 0
2
2


To find the zero input response, you will need the characteristic mode found earlier, H1e  t  H 2 e 2t .
Using the initial conditions given for y  0  and y  0  you can derive the input response yzi  t  as you
did before.
yzi  t    H1e  t  H 2e 2t 
yzi  0   6  H1  H 2


 H1  17, H 2  11
yzi  0   5   H1  2 H 2 

yzi  t   17et  11e2t
t 0
The total solution is now
y  t   yzs  t   yzi  t 
y  t    0.5 et  0.5 e3t   17et  11e 2t  t  0
y  t   17.5et  11e2t  0.5e 3t
t 0
Example 2
For this example we’ll add a second order derivative of the input and solve for the impulse response.
Since the order of the differentials are equal on the left and right, there will be a delta function present
in the impulse response.
d2
d
d2
d
d
y
t

3
y
t

2
y
t

3
x
t

x
t

2
x
t
where
y
0

6
and
y 0  5














dt 2
dt
dt 2
dt
dt
x  t   e 3t u  t 
Before we can obtain the systems impulse response, we must first solve for h0  t  .
d2
d
Q  D  2  3  2
dt
dt
d2 d
P  D  3 2   2
dt
dt
The equation for h0  t  is the same as Example 1,
 d2

d
 dt 2  3 dt  2 h0  t     t  t  0


which was found to be
h0  t   et  e2t
t 0
We can now find h  t  using
h  t   c0   t   P  D  h0  t 
 d2 d

 3  t   3 2   2  et  e2t  t  0
 dt dt

t
2 t
 4e  12e
t 0
Knowing the impulse response of the system, you can now use its convolution with the input x  t  to
find the zero-state response. The zero-input response will be the same as found in Example 1.