Differential Equations
Grinshpan
Principle of Superposition. Wronskian.
Linear homogeneous differential equations
x(n) (t) + pn−1 (t)x(n−1) (t) + · · · + p1 (t)x0 (t) + p0 (t)x(t) = 0
have a linear structure of the solution set. We will illustrate this for n = 2.
Let x = ϕ(t) and x = ψ(t) be solutions of the equation
(1)
ẍ(t) + P (t)ẋ(t) + Q(t)x(t) = 0.
on the interval α < t < β. Then for any constants a and b the linear combination
x = aϕ(t) + bψ(t) is also a solution of (1) on (α, β). This is an instance of the
superposition principle, and one says that solutions of (1) form a vector space.
Proof:
(aϕ + bψ)¨+ P (aϕ + bψ)˙+ Q(aϕ + bψ) = aϕ̈ + bψ̈ + P aϕ̇ + P bψ̇ + Qaϕ + Qbψ
= a(ϕ̈ + P ϕ̇ + Qϕ) + b(ψ̈ + P ψ̇ + Qψ)
= a0 + b0 = 0.
Thus any two solutions generate an infinite set of solutions. We would like this set
to include as many solutions as possible.
Let α < t < β and fix α < t0 < β. Consider the initial value problem


 ẍ(t) + P (t)ẋ(t) + Q(t)x(t) = 0
x(t0 ) = x0
(2)
.


ẋ(t0 ) = x1
Let us try to find constants a and b so that x = aϕ(t) + bψ(t) has given initial
conditions. We must have
(
aϕ(t0 ) + bψ(t0 ) = x0
.
aϕ̇(t0 ) + bψ̇(t0 ) = x1
Solving for a and b we find that if ϕ(t0 )ψ̇(t0 ) − ϕ̇(t0 )ψ(t0 ) 6= 0 then there is a
unique choice of constants:

x0 ψ̇(t0 ) − x1 ψ(t0 )



a =
ϕ(t0 )ψ̇(t0 ) − ϕ̇(t0 )ψ(t0 )
.

−x0 ϕ̇(t0 ) + x1 ϕ(t0 )


b=
ϕ(t0 )ψ̇(t0 ) − ϕ̇(t0 )ψ(t0 )
The function W (t) = ϕ(t)ψ̇(t) − ϕ̇(t)ψ(t) is called the Wronskian determinant of
ϕ and ψ. Symbolically,
ϕ ψ .
W =
ϕ̇ ψ̇ We have obtained the following conclusion.
2
Conclusion. Let ϕ(t) and ψ(t) be solutions of (1) on (α, β). If W (t0 ) 6= 0 then
every initial value problem (2) is solvable. In fact, (2) has a unique solution of the
form aϕ(t) + bψ(t).
Example. Let t > 0 and consider the initial value problem

2

 ẍ(t) − 2/t x(t) = 0
x(1) = 0
.


ẋ(1) = 2
By inspection we can check that ϕ(t) = t2 and ψ(t) = 1/t both satisfy
ẍ(t) − t22 x(t) = 0. However neither function has the right initial conditions:
ϕ(1) = 1, ϕ̇(1) = 2
and ψ(1) = 1, ψ̇(1) = −1.
The Wronskian of ϕ and ψ at t = 1 is
1 1
W (1) = 2 −1
= −3 6= 0.
Hence there are constants a, b such that x = at2 + b/t is a solution of our initial
value problem. A short computation gives
−2 2
2 1
2
x(t) =
t +
= (x2 − 1/t).
−3
−3 t
3