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Second Order Linear Differential Equations
We’ve looked at special cases of the differential equation
ay
′′
′
where a, b, and c are constants.
If we consider the more general homogeneous equations
y
′′
′
where p(t) and q(t) are functions of t, when can we be assured that a solution
exists?
Theorem. Let I be an interval that contains the point t 0 , and suppose
the functions p(t), q(t) and g(t) are continuous on I. Then there is
exactly one solution to the initial value problem
y
′′
′
- q(t)y = g(t), y(t 0 ) = y 0 , y
′ (t 0 ) = y
′ 0
valid in the interval I.
Example. Find the largest interval in which the solution to the initial value
problem
(cos t)y
′′
′
′ (0) = 3.
is certain to exist
Rewrite the differential equation in the form appearing in the theorem
y
′′
t
cos t
y
′
cos t
y =
cos t
The coefficient functions
t
cos t
cos t
cos t
are continuous in the interval (−π/ 2 , π/2). This interval contains t = 0.
Therefore, there is a unique solution to the IVP on the interval (−π/ 2 .π/2).
Anytime we have two solutions of a second order homogeneous linear dif-
ferential equation, their linear combination is also a solution.
This result is sometimes called the principle of superposition.
Theorem. If y 1 and y 2 are solutions to the differential equation
y
′′
′
then the linear combination c 1 y 1 + c 2 y 2 is also a solution for any choice
of constants c 1 , c 2.
The superposition principle results from the basic differentiation rules
d
dt
(y 1 (t) + y 2 (t)) =
d
dt
y 1 (t) +
d
dt
y 2 (t),
d
dt
(Cy(t)) = C
d
dt
y(t), C ∈ R.
Key Questions:
If we have two solutions y 1 and y 2 to the equation
y
′′
′
- Is there a choice of constants for which y(t) = c 1 y 1 (t) + c 2 y 2 (t) also
satisfies some arbitrary initial conditions
y(t 0 ) = y 0 , y
′ (t 0 ) = y
′ 0?
- Can we get all the solutions through some linear combination
c 1 y 1 + c 2 y 2 , c 1 , c 2 ∈ R?
To answer these questions, we need to define a few facts from linear algebra.
Our calculations above show that, so long as the denominators in the
expressions for c 1 , c 2 above are nonzero, we can always satisfy the initial
conditions.
Theorem. Suppose that y 1 and y 2 are two solutions to
y
′′
′
and that the Wronskian
W (y 1 , y 2 ) = y 1 y
′ 2 −^ y
′ 1 y^2
is not zero at the point t 0 where the initial conditions
y(t 0 ) = y 0 , y
′ (t 0 ) = y
′ 0
are assigned. Then there is a choice of the constants c 1 , c 2 for which
y = c 1 y 1 (t) + c 2 y 2 (t)
satisfies the differential equation and the initial condiditions.
This means that we can always find a solution to an IVP
y
′′
′
- q(t)y = 0, y(t 0 ) = y 0 , y
′ (t 0 ) = y
′ 0
with a linear combination of two particular solutions, y 1 (t) and y 2 (t), to the
differential equation, provided W (y 1 , y 2 )(t 0 ) 6 = 0.
In fact, as shown on the last page, the solution is
y(t) = c 1 y 1 (t) + c 2 y 2 (t),
where c 1 and c 2 are defined by
c 1 =
y 0 y 2 (t 0 )
y
′ 0 y
′ 2 (t^0 )
y 1 (t 0 ) y 2 (t 0 )
y
′ 1 (t^0 )^ y
′ 2 (t^0 )
, c 2 =
y 1 (t 0 ) y 0
y
′ 1 (t^0 )^ y
′ 0
y 1 (t 0 ) y 2 (t 0 )
y
′ 1 (t^0 )^ y
′ 2 (t^0 )
Theorem. If y 1 and y 2 are two solutions of the differential equation
y
′′
′
and if there is a point t 0 where the Wronskian of y 1 and y 2 is nonzero,
then the family of solutions
y = c 1 y 1 (t) + c 2 y 2 (t)
includes each solution to (1) for some choice of constants c 1 , c 2.
Proof. Suppose you have an arbitrary function θ(t) satisfying the differential
equation
y
′′
′
we need to see that
θ(t) = k 1 y 1 (t) + k 2 y 2 (t), for some constants k 1 , k 2.
If y 1 and y 2 are two particular solutions to (1) and t 0 is a point where the
Wronskian of y 1 and y 2 is nonzero, the theorem on the last page implies that
we can choose c 1 and c 1 so that a solution to the initial value problem
y
′′
′
- q(t)y = 0, y(t 0 ) = θ(t 0 ), y
′ (t 0 ) = θ
′ (t 0 ) (2)
is given by
y(t) = c 1 y 1 (t) + c 2 y 2 (t).
Therefore, the two functions
y(t) = c 1 y 1 (t) + c 2 y 2 (t), and θ(t)
are both solutions to the IVP (2). By the uniqueness of solutions to the
IVP (2) from the first theorem in today’s notes, θ and y are really the same
function.
That is,
θ(t) = y(t) = c 1 y 1 (t) + c 2 y 2 (t).
Example. We proved yesterday that, if r 1 and r 2 are two different, real roots
of the quadratic equation
ar
2
then, for any choice of constants c 1 and c 2 ,
y(t) = c 1 e
r 1 t
r 2 t
is a solution to the differential equation
ay
′′
′
Do the solutions e
r 1 t and e
r 2 t form a fundamental set of solutions for the
differential equation on the interval (−∞, ∞)?
That is, are there any solutions that are not of the form
y(t) = c 1 e
r 1 t
r 2 t
for some choice of constants c 1 and c 2?
The Wronskian of y 1 (t) = e
r 1 t and y 2 (t) = e
r 2 t equals
W (y 1 , y 2 ) =
e
r 1 t e
r 2 t
r 1 e
r 1 t r 2 e
r 2 t
= r 2 e
(r 1 +r 2 )t − r 1 e
(r 1 +r 2 )t
= (r 1 − r 2 )e
(r 1 +r 2 )t .
Since r 1 and r 2 are different, r 1 6 = r 2 , so that r 1 − r 2 6 = 0.
No matter which t we choose in the interval (−∞, ∞),
W (e
r 1 t , e
r 2 t ) 6 = 0.
By the last theorem above, this means that y 1 (t) = e
r 1 t and y 2 (t) = e
r 2 t
form a fundamental set of solutions.
In other words, every solution, y(t), to the differential equation
ay
′′
′
y(t) = c 1 e
r 1 t
r 2 t
for some choice of constants c 1 and c 2.
