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Note on Linear Differential Equations
Econ 204A - Prof. Bohn
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We will have to work with differential equations throughout this course. Differential equations – and
their discrete-time analogs: difference equations – are economically interesting because they link
levels to changes; we are often interested in linking the current situation or status of an economy to
changes that we are trying to predict or understand.
This note is about linear differential equations, linear relationships between a variable and its time-
derivative. The general specification is
(1)
dy ( t )
dt
= !( t ) " y ( t ) + x ( t )
The variable y = y ( t ) is a function of time, to be determined. The derivative dy/dt is also a function of
time, variously denoted y ( t ) ( t ) y '( t ) y ( t )
dt
dy
dt
d
= = =. The terms γ(t) and x(t) are known functions of
time, called the coefficients or forcing variables. Solving a differential equation means writing y(t) as
function of time that does not involve the derivative.
Categorizations:
- Fixed and variable coefficients: The solutions simplify if γ and x are constants, also called fixed
coefficients. Caution : Writers often suppress time-dependence when working with differential
equations. That is, equation (1) is often written more compactly as
(1’) y = "! y + x
Readers are expected to determine from the context if γ and/or x are constant or if they should be
treated as variables.
- Homogenous and non-homogenous equations: A differential equation is homogenous if there is no
additive part, i.e., if x ( t )! 0 for all t. Otherwise it is non-homogenous. Useful fact: The solution to a
non-homogenous equation is always the solution to the homogeneous part—omitting the x-part—plus
a function of time. In some applications, the non-homogenous part is not economically interesting and
one can simply examine the homogenous part.
- General and special solutions: Equation (1) is typically solved by a parametric class of functions,
which is called the general solution to (1). To pin down a unique y-function—a specific solution —we
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Disclaimer and request: The Note is meant as a summary and reference, not a self-contained text.
You should also read Barro/Sala-i-Martin’s appendix and (if you need it) consult a suitable math-for-
economists text (e.g., Chiang). For the benefit of other students, please let me know if sections are
unclear or you find glitches.
need additional pieces of information called boundary conditions. For linear differential equations, the
general solution is indexed by a single parameter (denoted A below). A single boundary condition is
sufficient to determine this parameter.
For this note, I will assume that the boundary condition is the y-value at a particular time t=t 0
. That is,
y ( t
0
) = y
0
is assumed known and provides the boundary condition. In applications, we often
normalize t 0
=0. I show general solutions if they are more compact than specific solutions.
- Stable solutions, explosive solutions, and degenerate cases: Depending on the parameters, solutions
will display different types of asymptotic behavior. Important characteristics are stability—
convergence to a finite limit value as t! ", and explosive behavior—divergence to infinity.
Sometimes special cases deserve attention.
Main Cases:
Four cases are worth distinguishing. I will state all solutions and then explain how they are obtained.
1a. Homogenous equation with fixed coefficient γ: y ( t )= "! y ( t )
.
General solution: y ( t ) = A! e
" t
Specific solution: y ( t ) = y ( t
0
)! e
" ( t # t
0
)
Alternate notations are y ( t ) = y
0
! e
" ( t # t
0
)
or y ( t ) = y ( t
0
)! exp{ "( t # t
0
The general solution highlights the exponential shape of the function, which is the key feature. If
γ<0. the solution converges to zero from any starting value. If γ>0, the solution diverges to plus
infinity from any positive starting value and to minus infinity from any negative starting value. If
γ=0, y(t)=y(t 0
) is constant; and if y(t 0
)=0, y(t)=0 for all γ.
1b. Homogenous equation with variable coefficient γ(t): y ( t )= "( t )! y ( t )
.
General solution: y ( t ) = A! e
" ( s ) ds
Specific solution: y ( t ) = y ( t
0
)! e
" ( s ) ds
t
0
t
The indefinite integrate in the general solution highlights the key feature, the integral over the
forcing function in the exponent. In the specific solution, the γ-function is integrated over time
from the boundary time t
0
to time t. (The integration index s has no significance.)
2a. Non-homogenous equation with fixed coefficients γ and x: y ( t )= "! y ( t )+ x
.
General solution: y ( t ) = A! e
" t
x
"
for "! 0 ;
Specific solution: y ( t ) = y ( t
0
)! e
" ( t # t
0
)
x
"
! [ 1 # e
" ( t # t
0
)
] for "! 0 ;
The solutions are explosive if γ>0 and stable if γ<0, as in the homogenous case. If γ<0, a nonzero
x-value implies a “displacement” of the limit value away from zero. Note that (–x/γ)>0 if x>0 and
γ<0, so the limit has the same sign as x. For the intuition, note that setting dy/dt=0 yields 0=γy+x,
Proof #2: For a more intuitive proof, note that a change divided by a level has the dimension of a
growth rate. In the homogenous fixed-coefficient case, the differential equation (1) can be written as
dy ( t )
dt
/ y ( t ) =!
It simply says that y(t) has a constant growth rate. Also note that growth rates are log-differences:
d
dt
ln y ( t ) =
1
y ( t )
d
dt
y ( t ),
using the chain rule of differentiation. Because logs and exponentials are inverse operations, it should
not surprise that the solutions involve exponentials. To prove case 1a, define z ( t ) = ln( y ( t )) and write
(1) as! =
d
dt
ln y ( t ) = dz / dt. A function with a constant derivative γ must be linear with slope γ. (Or
more generally, integrating over a derivative recovers the function. Integrating over a constant yields a
linear function.) Thus, z has the form z ( t ) = a +! " t for some constant a. Taking exponentials,
y ( t ) = e
z ( t )
= e
a +! " t
= A " e
! " t
with A = e
a
. This provides a constructive proof for case 1a. The other
cases are generalizations, as follows.
Proof #3: An intuition for a general proof derives from the product rule of differentiation applied to
functions of the form z ( t ) = y ( t )! e
"( t )
. The derivative of z has the format
(2) z '( t ) = y '( t )! e
"( t )
"( t )
= [ y '( t ) + "'( t )! y ( t )]! e
"( t )
.
The key insight is that the derivative of z involves a linear combination of the function y and its
derivative y’. The fundamental theorem of calculus says that integrating over a derivative recovers the
function. The term e
!( t )
is called the integration factor. The key idea is that, if one can regroup the y
and y’ terms from the differential equation (1) in the form (2), with z’ equal to something known, one
can recover the z-function through integration. That’s the main intuition—everything else is tedious
algebra.
Turning to the algebra, compare (1) and (2): To obtain the same linear combination of y and y’, we
need !'( t ) = " #( t )
. Integrating over Λ’, one finds that the integration factor must have the form
e
!( t )
= exp{"$ #( v ) dv }
. Because we have a t
0
- dated boundary information, this is written most
conveniently as e
!( t )
= exp{" #( v ) dv
t
0
t
0
} with some integration constant a
0
and integration
starting at time t 0
. Writing (1) as y '( t )! "( t ) # y ( t ) = x ( t ) and multiplying by e
!( t )
, we find
z '( t ) = [ y '( t )! "( t ) # y ( t )]# e
$( t )
= x ( t ) # e
$( t )
.
This can be integrated to obtain z ( t ) = y ( t )! e
"( t )
= x ( v )! e
"( v )
t
0
t
dv + a
1
for some integration
constant a 1
. Then divide by e
!( t )
to obtain y ( t ) = a
1
! e
"#( t )
#( v )
0
t
$ dv! e
"#( t )
. Finally, note
that e
!"( t )
= exp{ #( v ) dv! a
t 0
0
t
$ } and e
!( v )"!( t )
= exp{ #( s ) ds
v
t
$ } are identical to the
corresponding terms in 2b. Comparing like terms, one may interpret y
0
= a
1
! e
" a
0
as combination of
integration constants. This proves case 2b. The others case are specializations.
Optional Exercise: Prove cases 1a, 1b, and 2a directly by defining suitable integration factors and
going through the same steps as above.
Final note on variable coefficients: γ and x are both fixed in case 2a and both variable in case 2b. An
intermediate case would be a constant γ and variable x : y! ( t )= "! y ( t )+ x ( t ). In this case, the
solution to 2b would simplify to y ( t ) = y
0
! e
" ( t # t
0
)
" ( t # v )
t
0
t
dv. If x is also constant, it can
be taken out of the integral. Then for "! 0 , the second integral reduces to
x ( v )
t
0
t
e
" ( t # v )
dv = x $ e
" ( t # v )
t
0
t
dv =
x
"
[ e
" ( t # t
0
)
1 ], This integration explains why the ratio x/γ
appears in case 2a. The ‘A’ in the general solution 2a corresponds to y
0
intermediate case reduces to y ( t ) = y
0
t
0
t
dv , or a linear function for constant x.
