Wouldn’t it be nice if there was an easy way to determine linear independence or dependence of a set of functions?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
Well, with the Wronskian, it’s easy! And that’s only part of what the Wronskian determinant can do!
It can also help us solve homogeneous and nonhomogeneous second and higher-order differential equations. Cool, right?
Okay, so let’s look at the Wronskian and its special properties.
Definition and Properties of the Wronskian
Suppose a set of functions \(f_{1}(x), f_{2}(x), \ldots, f_{n}(x)\) possess derivatives of order \(n-1\). Then the Wronskian determinant, where the primes denote derivatives, is as follows:
\begin{align*}
W\left(f_{1}, f_{2}, \ldots, f_{n}\right)=\left|\begin{array}{cccc}
f_{1} & f_{2} & \cdots & f_{n} \\
f_{1}^{\prime} & f_{2}^{\prime} & \cdots & f_{n}^{\prime} \\
\vdots & \vdots & \ddots & \vdots \\
f_{1}^{(n-1)} & f_{2}^{(n-1)} & \cdots & f_{n}^{(n-1)}
\end{array}\right|
\end{align*}
But how does the determinant prove linearly independent solutions?
Linear Independence and the Wronskian
Well, the solutions of a homogeneous nth-order differential equation on some interval \(I\) are linearly independent if and only if \(W\left(f_{1}, f_{2}, \ldots, f_{n}\right) \neq 0\) for every \(\mathrm{x}\) in the interval.
Examples: Using the Wronskian Determinant
Let’s see this powerful determinant in action with a few examples.
Example 1: Two Functions
Set up the Wronskian for the set of functions \(y_{1}=4 x, y_{2}=e^{x}\) and determine if the set is linearly independent or dependent.
First, we will construct our determinant. Because we are only given two functions, our determinant will be \(2 \times 2\)
\begin{align*}
W=\left|\begin{array}{cc}
y_{1} & y_{2} \\
\frac{d y_{1}}{d x} & \frac{d y_{2}}{d x}
\end{array}\right|=\left|\begin{array}{cc}
4 x & e^{x} \\
4 & e^{x}
\end{array}\right|
\end{align*}
Next, let’s remind ourselves how to calculate the determinant of a \(2 \times 2\) matrix.
\begin{equation}
\operatorname{det}\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c
\end{equation}
Now, it’s time to find the Wronskian determinant for our functions.
\begin{align*}
W=\left|\begin{array}{cc}
4 x & e^{x} \\
4 & e^{x}
\end{array}\right|=(4 x)\left(e^{x}\right)-4\left(e^{x}\right)=4 e^{x}(x-1)
\end{align*}
Lastly, we notice that our Wronskian does not equal zero. This means that our set of functions are linearly independent.
Nice!
Now, let’s look at a set of three functions.
Example 2: Three Functions
Set up the Wronskian for the set of functions \(y_{1}=x, y_{2}=x e^{x}\), and \(y_{3}=3 x\) and determine if the set is linearly independent or dependent.
First, we will construct our determinant. Because we are only given two functions, our determinant will be \(3 \times 3\)
\begin{align*}
W=\left|\begin{array}{ccc}
y_{1} & y_{2} & y_{3} \\
\frac{d y_{1}}{d x} & \frac{d y_{2}}{d x} & \frac{d y_{3}}{d x} \\
\frac{d^{2} y_{1}}{d x^{2}} & \frac{d^{2} y_{2}}{d x^{2}} & \frac{d^{2} y_{3}}{d x^{2}}
\end{array}\right|=\left|\begin{array}{ccc}
x & x e^{x} & 3 x \\
1 & x e^{x}+e^{x} & 3 \\
0 & x e^{x}+2 e^{x} & 0
\end{array}\right|
\end{align*}
Next, we will calculate the determinant using expansion by minors, sometimes called expansion by cofactors.
\begin{align*}
W=\left|\begin{array}{ccc}
x & x e^{x} & 3 x \\
1 & x e^{x}+e^{x} & 3 \\
0 & x e^{x}+2 e^{x} & 0
\end{array}\right|=0
\end{align*}
And what do we notice? Our Wronskian equals zero. This means that our set of functions are linearly dependent. So, together we will begin our lesson by understanding differential operators and their notation and discuss whether or not it is possible to write a useful expression for a solution to a second-order linear differential equation (i.e., existence and uniqueness theorem).
Wronskian Determinant in Differential Equations
This prompts us to revisit the Principle of Superposition, which states that if we have two known solutions to the differential equation, then for any constants, the linear combination of both solutions is also a solution. But this begs us to ask, “When do we have a unique solution and how do we solve for these arbitrary constants?”
To answer that question, we must:
- Determine linear independence
- Investigate the Wronskian determinant and Abel’s Theorem
What is always important to keep in mind is that the Wronskian helps us to develop a condition that will allow us to determine when we can solve for the arbitrary constants.
Thus, it answers our question as to whether or not our two known solutions form a fundamental set of solutions and paves the way for our future lesson involving variation of parameters and how to solve nonhomogeneous DEs.
Alright, let’s get after it!
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