MATH11219
WEEK 10 CHAPTER 20 Blocks 1 -3
CLASSIFYING DIFFERENTIAL EQUATIONS
df
d2 f
An equation in terms of
,
, .... etc is a differential equation.
dx
dx 2
The order of a differential equation is related to the highest order of the derivative in the equation.
The independent variable is the variable with respect to which the differential is taken.
The dependent variable is the variable which is differentiated.
dx
eg in
x is the dependent variable and t is the independent variable
dt
Linear Equations
Linear differential equations are those in which the dependent variable or variables and their derivatives do not
occur as products, raised to powers or in non-linear functions (eg trig functions, exponential functions, etc)
In a constant-coefficient linear differential equation, the coefficients of the dependent variable and its derivatives
are constants.
Reference: p911 - 913
SOLVING DIFFERENTIAL EQUATIONS
To show that a function is a solution of a differential equation, find the appropriate derivatives and substitute
into the differential equation to prove that LHE = RHE.
Reference: Examples 1.6 - 1.8 on p916
From our previous work with derivatives and integrals, we know that:
If the rate of change is a function of the independent variable,
dx
 f (t )
dt
then
This gives us the general solution
x 

f (t ) dt
x = F(t) + C
where C is an arbitrary constant
If we are given additional information, we are able to obtain the value of the arbitrary constant and hence obtain
the particular solution.
If we are given a second order differential, we need to integrate twice. There will be an arbitrary constant at each
stage of the integration when finding the general solution. If information is provided for a particular solution,
then each arbitrary constant needs to be calculated at each stage of the integration.
Reference: Examples 1.9 - 1.11 on p919 - 921
If the rate of change is a function of the dependent variable such that
then
dx
 kx
dt
dx = kx dt
1
dx  k dt
x
1
 x dx   k dt
ln x  kt  C
x  e
k t C
x  e kt . e C
x  A e kt
Hence,
dx
 kx
dt
then
x  Ae kt
where A is the initial value.
If the rate of change is a function of the both variables, we rearrange the differential equation with all terms
containing the dependent variable on the left-hand side (LHS) of the equation and those only containing the
independent variables and constant terms on the right-hand side (RHS).
dy
 f ( x )g ( y )
dx
1
dy  f ( x ) dx
g( y )
1
 g( y ) dy   f ( x ) dx
Note: 1. Even though we integrate each side, we only need to add one arbitrary constant, usually on the RHS.
2. We usually rearrange the function to give the dependent variable as the subject.
Reference: Examples 2.1 – 2.7 on p924 - 929
FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
The standard form for a first-order linear differential equation is
dy
 P( x ) y  Q( x )
dx
To solve a differential equation of this form we need to determine the integrating factor.
If  = e kx is the integrating factor, then  = e kx = e 
so that
y =
  Q( x )
dx
P( x ) dx
PROOF:
dy
 P( x ) . y  Q( x )
dx
If
we multiply throughout by the integrating factor
so that
e k( x )
 = e k( x )  e 
P( x ) dx
dy
 e k ( x ) P( x ) . y  e k ( x ) Q( x )
dx
The left hand side is the reverse of the product rule, so we obtain
d k( x )
e
.y
dx


y . e k( x )

y
e k ( x ) . Q( x )
e
k( x )
 e  k( x )
. Q( x ) dx
 e
k( x )
. Q( x ) dx

Reference: Process as explained on p933 - 935 and Examples 3.1 – 3.6 on p9931 - 937