ODE Lecture Notes
Section 3.2
Page 1 of 6
Section 3.2: Solutions of Linear Homogeneous Equations; the Wronskian
Big Idea: We can be certain that any two solutions we find of a homogeneous linear secondorder differential equation are the only two solutions possible if a quantity called the Wronskian
is non-zero on the interval of the existence of the solution.
Big Skill: You should be able to compute the Wronskian and use it to determine if you have
found a fundamental set of solutions.
Note on how to get ODE Architect to numerically solve a second-order differential
equation:
To graph the solution to a second-order system in ODEA, you have to pull the trick of creating a
"new" dependent variable to represent the first derivative, which results in a system of first-order
differential equations.
So, for example, to numerically solve
y'' + y' - 2y = 0, y(0) = 1, y(0) = 0
first solve for y'': y'' = -y' + 2y
Then let v = y' and replace all occurrences of y' with v.
That results in the system below, which can be solved numerically in ODE Architect:
y' = v
v' = -v + 2*y
ODE Lecture Notes
Section 3.2
Page 2 of 6
Linear Second-Order Differential Operator L:
L       p   q … L  D2  pD  q
We will use this notation to examine initial value problems of the form
L  y   0, y  t0   y0 , y t0   y0
Theorem 3.2.1: Existence and Uniqueness Theorem
(for Linear Second-Order Differential Equations)
For the initial value problem
y  p  t  y  q  t  y  g t  , y t0   y0 , y t0   y0
where p, q, and g are continuous on some open interval I containing the point t0, then there is
exactly one solution y    t  of this problem, and the solution exists throughout the interval I.
Note the theorem asserts three things:
 A solution exists.
 The solution is unique.
 The solution  is defined throughout I, and is thus at least twice differentiable there.
Practice:
1. Find the longest interval in which the solution of the IVP is certain to exist for
1
cos  t  y  y  9  t 2 y  0, y  2   1, y  2   0 . State the linear operator associated
t
with this differential equation.
ODE Lecture Notes
Section 3.2
Page 3 of 6
Theorem 3.2.2: Principle of Superposition
If y1 and y2 are solutions of the differential equation
L  y   y  p  t  y  q  t  y  0 ,
then the linear combination c1y1 + c2y2 is a solution for any values of the constants c1 and c2.
Proof:
Next issue: are all solutions of L  y   y  p  t  y  q  t  y  0 included in y  c1 y1  t   c2 y2  t  ,
or are there other solutions?
Begin examination by looking at what is required of c1 and c2 to satisfy the initial conditions
y  t0   y0 , y  t0   y0 . They must obey the system of equations:
c1 y1  t0   c2 y2  t0   y0

c1 y1  t0   c2 y2  t0   y0
By Cramer’s Rule (from algebra), the solution of this system is:
y0 y 2  t 0 
y1  t0  y0
y0 y2  t0 
y1  t0  y0
c1 
, c2 
y1  t0  y2  t0 
y1  t0  y2  t0 
y1  t0  y2  t0 
y1  t0  y2  t0 
Thus, the system has no solution if the determinant in the denominator is zero. This determinant
gets a special name, the Wronskian:
y  t  y2  t 0 
W 1 0
 y1  t0  y2  t0   y2  t0  y1  t0 
y1  t0  y2  t0 
An alternative way to represent that the Wronskian depends on the functions y1 and y2 evaluated
at t0 is: W  y1 , y2  t0  … instead of just W.
ODE Lecture Notes
Section 3.2
Page 4 of 6
Practice:
2. Compute the Wronskian of the solutions to the differential equation
y  12 y  20 y  0
Theorem 3.2.3:
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
and the initial conditions y  t0   y0 , y  t0   y0 must be satisfied by y, then it is always possible
to choose the constants c1 and c2 so that
y  c1 y1  t   c2 y2  t 
satisfies this initial value problem if and only if the Wronskian W  y1 y2  y2 y1 is not zero at t0.
Theorem 3.2.4:
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
then the family of solutions
y  c1 y1  t   c2 y2  t 
with arbitrary constants c1 and c2 includes every solution of the differential equation if and only
if there is a point t0 where the Wronskian W  y1 y2  y2 y1 is not zero.
Notes:
 This is like when we solved y  4 y  0 subject to y  0  1, y  0  1 and got two very

different looking solutions y  t   0.25e2t  0.75e2t and y  t   cosh  2t   0.5sinh  2t  ,
but that represented the same function.
y  c1 y1  t   c2 y2  t  is called the general solution of L  y   y  p  t  y  q  t  y  0 .

y1 and y2 are said to form a fundamental set of solutions.
ODE Lecture Notes

Section 3.2
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To find the general solution, and thus all solutions of L  y   y  p  t  y  q  t  y  0 , we
need only find two solutions of the given equation whose Wronskian is non-zero.
Practice:
3. State whether the solutions found in problem #2 form a fundamental set of solutions.
4. Determine whether y1  t   t1/2 and y2  t   t 1 for a fundamental set of solutions of
2t 2 y  3ty  y  0 for t > 0.
Theorem 3.2.5:
If the differential equation
L  y   y  p  t  y  q  t  y  0
has coefficients p and q that are continuous on some open interval I, and if t0  I, and if y1 is a
solution that satisfies the initial conditions
y1  t0   1, y1  t0   0
and if y2 is a solution that satisfies the initial conditions
y2  t0   0, y2  t0   1
then y1 and y2 form a fundamental set of solutions of the equation.
Proof:
ODE Lecture Notes
Section 3.2
Page 6 of 6
Practice:
5. Find a fundamental set of solutions as specified by theorem 3.2.5 for y  4 y  0
Theorem 3.2.6: (Abel’s Theorem)
If y1 and y2 are two solutions of the differential equation
L  y   y  p  t  y  q  t  y  0
With the coefficients p and q continuous on an open interval I, then the Wronskian W  y1 , y2  t 
is given by:
W  y1 , y2  t   c exp    p  t  dt 
where c is a certain constant that depends on y1 and y2, but not on t. Further, W  y1 , y2  t  either
is zero for all t  I (if c = 0), or else is never zero in I (if c  0). The constant c can be
determined by evaluating W at some convenient point.
Practice:
6. Compute the Wronskian for 2t 2 y  3ty  y  0 , and compare to practice #4.