EXAMPLE: THE WRONSKIAN DETERMINANT OF A SECOND-ORDER, LINEAR HOMOGENEOUS DIFFERENTIAL EQUATION

110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN

Given a second order, linear, homogeneous differential equation

y′′ + p(t)y′ + q(t)y = 0,

where both p(t) and q(t) are continuous on some open t-interval I, and two solutions y_{1}(t) and y_{2}(t), one can form a fundamental set of solutions as the linear combination of these two

y(t) = c_{1}y_{1}(t) + c_{2}y_{2}(t)

ONLY under the condition that the Wronskian determinant

W (y_{1}, y_{2})(t) = | \begin{matrix}y_{1}(t) & y_{2}(t) \\ y′_{1}(t) & y′_{2}(t) \end{matrix} | ≠ 0

for all t ∈ I. This condition implies that the two differentiable functions y_{1}(t) and y_{2}(t) are independent; that there does not exist two constants k_{1}, k_{2} ∈ ℝ, not both equal to 0, where k_{1}y_{1}(t) + k_{2}y_{2}(t) = 0 on I. Then we know that ALL solutions to the ODE will be a linear combination of y_{1}(t) and y_{2}(t).

We can actually take this further. It turns out that in this situation, and on the interval I, either of two things can happen:

• The Wronskian is always 0 on I (we say W (y_{1}, y_{1}) is identically 0 on I, or W (y_{1}, y_{2}) ≡ 0 on I), or

• the Wronskian is NEVER 0 on I.

Why? Because of the following:

Theorem. If y_{1}(t) and y_{2}(t) are two solutions to the ODE y′′ + p(t)y′ + q(t)y = 0, where p(t) and q(t) are continuous on some open t-interval I, then

W (y_{1}, y_{2})(t) = Ce^{− ∫ p(t) dt}

where C depends on the choice of y_{1} and y_{2}, but not on t.

First, some notes:

• If y_{1}(t) and y_{2}(t) are linearly dependent, then C = 0.

• If y_{1}(t) and y_{2}(t) are linearly independent, then W (y_{1}, y_{2}) ≠ 0 on ALL of I.

• Linear independence and non-zero Wronskian are the same thing for solutions to these ODEs.

Proof. Since y_{1} and y_{1} both solve the ODE, we have

y′′_{1} + p(t)y′_{1} + q(t)y_{1} = 0
y′′_{2} + p(t)y′_{2} + q(t)y_{2} = 0.

Here, multiply the first equation by −y_{2} and the second by y_{1} and then add them together (this will eliminate the coefficient q(t))

\underbrace{(y_{1}y′′_{2} − y_{2}y′′_{1})}_{W′(y_{1},y_{2})} +p(t) \underbrace{(y_{1}y′_{2} − y_{2}y′_{1})}_{W (y_{1},y_{2})} = 0,

Which leads to a first order differential equation whose variable IS the Wronskian determinant itself (really, the Wronskian is a function of t)

W′ + p(t)W = 0.

This first order ODE is both linear and separable, and by separation of variables, we get

\frac{W′}{W} = −p(t) ⟹ ln |W | = − ∫ p(t) dt + K ⟹ Ce^{− ∫ p(t) dt}.

□

Hence by the notes above just before the proof, either C = 0, and the Wronskian is always 0, and the two solutions are linearly dependent, or C ≠ 0, and the Wronskian is NEVER 0, and the two solutions are linearly independent.