Download Lecture Notes 2d order homogeneous DEs with constant coefficients

We have considered so far first order differential equations both linear and non-linear.
Tonight we begin considering differential equations of higher order.
dv
m
f( t ) which describes the velocity.
dt
As an example consider the example of motion:
Once we solve this DE for v(t) we use the fact that dx/dt = v(t) to solve for position.
But
dv
d
2
x( t ) so it is natural to ask can we solve for the position directly without having to first solve
dt d t 2
for the velocity via the second order linear differential equation :
2
d
m
x( t ) f( t ) ?
d t2
The answer is yes.
Tonight we are going to concentrate on the general second order homogeneous linear differential
equation with constant coefficients : a 
2
d y
b
dy
cy 0 .
dx
2
dx
We will then consider in great detail the application of this DE to the motion of a mass on a spring.
However, before we begin we need to discuss some basic notions concerning nth order linear
differential equations in general.
The general nth order homogeneous linear differential equation is an equation of the form:
a n( x) 
n
d y
an
n
dx
n
d y
d
n
1
n
dx
n
an
( x) 
1
y
a 1( x) 
...................
1
d
n
1
y
dy
dx
n
...................
1
a 1( x) 
a 0( x)  y 0
dy
a 0( x)  y f( x) then it is not homogeneous
dx
dx
dx
(because of f(x) on the right hand side).
If any of the derivatives are raised to a power or if there are any terms which are the products of the
If
a n( x) 
1
( x) 
various derivatives and or y it is not linear. For example : y 
2
d y
dy
2
dx
dx
2
0 is not linear for two reasons,
what are they?
For the IVP : a n( x) 
n
d y
n
dx
an
1
( x) 
d
n
1
n
dx
y
1
...................
a 1( x) 
dy
dx
a 0( x)  y 0
with initial conditions for y(0),y ' (0)................yn-1(0) has a unique solution of the form :
y( x) A 1 y 1( x) A 2 y 2( x) ......................................... A n  y n( x)
where the y i (x) am linearly independent solutions to the DE and the Ai are constants determined from
the initial conditions.
The question is what do we mean by linearly independent functions ?
Definition
Two functions are linearly dependent if there exist non-zero constants c1 and c2 such
that
c1f1(x) +c2f2(x) = 0 i.e. f1(x) is a scalar multiple of f2(x). Functions which are not linearly dependent are
called linearly independent.
Eg: sin(x) and cos(x) are linearly independent.
Suppose not then
Then
c2
sin ( x)
c1
c1sin(x) +c2cos(x) = 0 for some c1 and c2.
 cos ( x) but if x = /2 we obtain 1 = 0. therefore sin(x) and cos(x) must be
linearly independent.
Eg : ex and ex are linearly independent unless  = .
Suppose c1ex + c2ex = 0 for some c1 and c2. Then ex
= c ex where c =
c2
.
c1
taking the natural log of both sides x = ln(c) + x
x(- ) = ln(c)
if x = 1 then (- ) = ln(c)
if x = -1 then (- ) = - ln(c)
adding the two equations we have  -  = 0 or  = .
A little later we'll develop a much more direct way of establishing the linear independence of any number
of functions (the Wronskian) but as regards second order linear differential equations with constant
coefficients we have everything we need as regards linear independence of functions.
Let's return to the general second order homogeneous linear differential equation with constant
coefficients :
a
(I)
2
d y
2
dx
b
dy
cy 0 .
dx
We start by looking for solutions of the type y (x) = e x .
Then
dy
dx
 e
x
2
and
d y
2 x
 e .
2
dx
Putting these into our DE we obtain :
2 x
x
x
a   e
b   e
c e 0
Dividing through by e
x
we obtain : a  
2
b 
c 0. This is called the characteristic polynomial or the
characteristic equation of the differential equation.
The roots of this equation lead us to The general solution to (I) above.
This is of course just a quadratic whose solutions are given by the quadratic formula :
 1
b
b
2
4 ac
b
and  2
b
2
4 ac
2a
2a
There are 3 possibilities depending on the number and type of roots.
Case 1 Real Distinct Roots b
Then  1
y( x) A  e
b
 1 x
b
2
2a
B e
4 ac
2
4 ac  0
b
and  2
b
4 ac
and the general solution to (I) is :
2a
 2 x
2
As an example consider the IVP :
2
d y
5
dy
6  y 0 with y(0) = 1 and y ' (0) = 0
dx
2
dx
Here note that a = 1, b = - 5, and c = 6 so the characteristic equation is :
2 - 5 + 6 = 0 .
 - 3) ( -2) = 0
 = 3 and  = 2
Therefore the general solution is : y(x) = A e 2x + B e 3x.
Applying the initial condition y(0) = 1 we obtain
A+B=1
Applying the initial condition y'(0) = 0 we obtain
2A + 3 B = 0.
Solving this system we obtain : A = 3 and B = - 2.
2
Therefore the solution to the IVP :
d y
5
dy
6  y 0 with y(0) = 1 and y ' (0) = 0 is y(x) = 3 e 2x - 2
dx
2
dx
e 3x.
Case 2 Complex Roots b
2
4 ac  0
Here we'll use Euler's Identity : e i = cos() + i sin() where i =
In this case both possibilities
 1
b
i
b
2
2a
result as you should verify so let's work with 1.
4 ac
and  2
1
b
i
b
2
2a
4 ac
yield the same
For ease of calculation let's write 1 =  + i where 
Then the solution to a 
2
d y
b
dy
b
b
and 
2a
2
4 ac
.
2a
is y(x) = e ( + i) = e ei = e ( cos() + isin() ) .
cy 0
dx
2
dx
The only way this can solve the DE is if the real part of the solution, e  cos(), and the imaginary part
of the solution, e sin(), independently are solutions.
Therefore the general solution is : y(x) = e  ( Acos() +Bsin() ) .
2
Example Solve
d y
dy
2
dx
dx
2 y 0 with y(0) = 1 and y ' (0) = -1 .
Here the characteristic equation is : 2 +  + 2 = 0 .
The solutions are 
1
i
2
7
and
1

2
7
i
2
2
x
The general solution to the differential equation is : y( x) e  A  cos
2
7
x
2
B sin
7
x
2
Now we'll apply the initial conditions:
y( 0) 1 A
1
1 
1 1 
y ' (0) = -1 =
A
7B =
7B
2
2
2 2
1
Therefore A =1 B =
7
7
2
The solution to the IVP
d y
dy
2
dx
dx
2 y 0 with y(0) = 1 and y ' (0) = -1 is :
x
y( x) e
2
 cos
7
x
2
1
7
7  sin
7
x
2
Case 3 Repeated Real Roots b
b
In this case 1 = 2 =
.
2a
2
4 ac 0
b
2
Therefore one solution to the DE a
d y
2
dx
b
 dy
dx
cy 0 is y = e
2 a
. However, we know we need two
independent solutions.
In this case it turns out the general solution is y e
(Convince yourself that e
 x
and x e
 x
 x 
(A
B x) where the second solution is y e
are indeed linearly independent).
 x 
B x.
The reason for this deals with obtaining the general solution to a second order DE from a known solution
by a method called reduction of order i.e. :
If y1(x) is a solution to a second order DE then we can obtain the general solution by setting y(x) =
y1(x)v(x)
and solving for v(x).
Let's illustrate this for the case where we have a repeated root in the characteristic equation.
2
d y
Eg Solve
2
dy
1 0 .
dx
2
dx
2
The characteristic equation is 2 -2 + 1 = (  1) = 0 . Here we have the double root  = 1.
Therefore y = ex is a solution. We claim the general solution is y(x) = ex (A +Bx) is the general solution.
To obtain the general solution let y = ex v(x).
Then
dy x 
e v
dx
2
d y
x dv
e 
dx
x
e v
2
dx
(Verify this )
x dv
2e 
dx
2
x d v
e 
2
dx
2
d y
Putting these results in
2
dy
dx
x
e v
x dv
2 e 
dx
2
x d v
e 
2
dx
1 0 we obtain :
dx
2
x
2 e  v
x dv
e 
dx
x
e  v( x) 0
2
x d v
0
Which simplifies to : e 
2
dx
2
Which simplifies further to
d v
0
2
dx
Therefore
dv
B and
v Bx A
dx
x
x
Therefore y( x) e  v e  ( A Bx) as advertised.
The point is we don't have to go through this process each time : if there is a single root,  to the
characteristic equation then the general solution to a 
2
d y
2
dx
Differential Operator Notation
b
dy
dx
cy 0 is y(x) = ex ( A+Bx ) .
For the differential equation a 
2
d y
b
dy
cy in particular and for higher order differential equations
dx
2
dx
in general
obtaining the characteristic equation is made easier if we use what is called differential operator notation.
Define D =
d
dy
then
dx
Similarly D2 =
d
2
2
then
2
And in general Dn =
2
2
dx
= D2 y
dx
d
n
d y
n
n
dx
d y
d y
2
dx
Therefore a 
can be written as simply Dy. D is called a differential operator.
dx
b
dy
n
= Dn y.
dx
cy = 0 becomes (aD2 + bD +c)y = 0 . Compare this with the characteristic
dx
equation
a2 + b + c = 0
The point being that if an nth order differential equation with constant coefficients is written using
differential
operators then the characteristic equation is obtained simply by replacing D with .
As an example consider the 3d order eqn : 7 
3
d y
3
15 
2
d y
2
dx
dx
3
2
(7 D - 15D +4) y = 0 therefore the characteristic eqn is :
73 -152 + 4 = 0.
4 y 0 . Written using operator notation: