Solving Differential Equations
Using the Zero-State and Zero-Input Methodology
We’ll limit this discussion to second order differential equations of the form
D 2 y  t   a Dy  t   b y  t   x  t 
where D 
d
dt
y (0)  ?
y '(0)  ?
(1.1)
There are two forces at work in this equation. One is the characteristic modes of the system (the
homogeneous or general solution) and the other is the input or forcing function, x  t  , that will
drive a response in synch with it and is known as the particular solution.
General Solution Form
The General Solution is made from the characteristic modes of the system. These modes exist
even when there is no input, which reduces differential equation (1.1) to
(1.2)
D2 y  t   a D y  t   b y  t   0
Assuming a solution of the form y  t   e t , after substituting into Eqn. (1.2) and dividing out
the e  t we have
(1.3)
 2  a  b  0
This is known as the Characteristic Equation whose solution is
(1.4)

a  a 2  4b
2
There are three distinct modes depending on the value of  as shown in the table below:
Case
Roots
a 2  4b  0
Distinct Real
Roots: 1 , 2
e1t , e2t
Real Double
a  4b  0 Roots:  =  a
2
a  4b  0
2
General Solution, yg  t 
Modes
1
2
e at / 2 , t e at / 2
Complex Conjugate e at /2 cos  t
1   12 a  j
 at /2
2   12 a  j
e
yg  t   H1e1t  H 2e2t
yg  t    H1  H 2t  eat /2
yg  t   e at /2  H1 cos t  H 2 sin t 
sin  t
Now that we’ve found the General Solution’s form the next step is to find the Particular
Solution.
Particular Solution
The Particular Solution depends on the forcing function that is driving the system, e.g., the input
x  t  . Given x  t  , use the table below to determine the form of y p  t  .
x(t)
yp(t)
ke t
Ce  t
kt n (n  0, 1, 2, ...)
K nt n  K n 1t n 1  ...  K1t  K 0
k sin  t 

k cos  t 
A cos t  B sin t
k e t sin t 

k e t cos t 
e t  A cos  t  B sin  t 
Modification: If yp has same exact form as the yg , the General
Solution, then multiply by t.
Now that you know the form of the Particular Solution, substitute this into Eqn. (1.1) to find the
undetermined coefficients.
Example Step 1 - Finding the General Solution’s Form and the Particular
Solution
D 2 y  t   3 D y  t   2 y  t   e 3t u  t  where y  0   0 and
d
y 0  1
dt
The Characteristic equation can be found by setting the equation to zero and then substituting e  t into
the equation. However, by looking at the order of the derivative, you’ll note that it is the same as the
power of  in the characteristic equation allowing us to immediately write
2  3  2  0
Solving the characteristic equation gives
1  1
2  2
The form of the General Solution can be written as
yg  t   H1et  H 2e2t
Turning now to the Particular Solution, note from the table that given an exponential input, the
form of the Particular Solution is
y p  t   Ce3t
Substituting this into the original differential equation yields
C  3 e3t  3 C  3 e3t  2 Ce3t  e3t
2
Solving for C gives C 
1
which in turn provides the Particular Solution
2
y p t  
1 3 t
e u t 
2
To find the total solution we’ll again breakdown the process into two parts; the zero-state
solution and the zero-input solution. The Zero-State solution makes the assumption that the
system is at rest (all initial conditions are zero) while the Zero-Input solution assumes that there
is no input but the system may not be at rest. We’ll tackle the Zero-State solution first.
Zero-State Solution
The Zero-State form is made up of both particular and general solutions
yzs (t )  General Form  Particular Form
(1.5)
where all initial conditions are set to zero. Since we’re limiting ourselves to second order
equations we need two initial conditions:
yzi (0)  0
d
yzi (0)  0
dt
(1.6)
Returning to our example…
Example Step 2 - Finding the Zero-State Solution
y zs (t )  General Form  Particular Form
1

  H1e  t  H 2 e 2t    e 3t 
2

Using the initial conditions gives
1
y zs (0)  0   H1  H 2    
2
d
 3 
y zs (0)  0    H1  2 H 2    
dt
 2 
Solving gives H1  1
2
and H 2  1. The Zero-State solution is then
1
1

yzs (t )   et  e2t  e3t  u  t 
2
2

The final step is to calculate the Zero-Input solution.
Zero-Input Solution
The Zero-Input solution is based on the General Solution’s form only.
(1.7)
yzi (t )  General Form
All that is left is to determine the unknown coefficients using the initial conditions.
Example Step 3 - Finding the Zero-Input Solution
yzi (t )  General Form
 H1et  H 2e2t
Using the given initial conditions we have
yzi (0)  0   H1  H 2 
d
yzi (0)  1    H1  2 H 2 
dt
Solving gives H1  1 and H 2  1. The Zero-Input solution is then
yzi (t )   et  e2t  u  t 
Total Solution
The total solution is simply the combination of the zero-state and zero-input solutions
y  t   yzs (t )  yzi (t )
(1.8)
So, for our example…
Example Step 4 – The Total Solution
y  t   y zs (t )  y zi (t )
1
1

  e  t  e 2t  e 3t    e  t  e 2t 
2
2

3
1
 e  t  2e 2t  e 3t where t  0.
2
2
Wrap-Up
This methodology does have its limitations. For example, we’ve made an assumption that the initial
conditions just before t  0 , notated as 0  , are the same as just after t  0 , notated as 0  . For this to
be true there cannot be a delta function at the origin. What this means for us is that the input cannot
contain a derivative:
D2 y t   a D y t   b y t  
d
x t 
dt
Note that in the equation above, if x  t   u  t  , there would be a delta function at the origin since the
derivative of a step function is a delta function. To solve this differential equation we must approach it
differently which we will in another lesson.