8.3: The Wronskian
Suppose that having determined that two solutions of Equation \ref{8.4} are \(x=X_{1}(t)\) and \(x=X_{2}(t)\) , we attempt to write the general solution to Equation \ref{8.4} as Equation \ref{8.5}. We must then ask whether this general solution will be able to satisfy two initial conditions given by
\[x\left(t_{0}\right)=x_{0}, \quad \dot{x}\left(t_{0}\right)=u_{0}, \nonumber \]
for any initial time \(t_{0}\) , and initial values \(x_{0}\) and \(u_{0}\) . Applying these initial conditions to \(Equation \ref{8.5}\) , we obtain
\[\begin{aligned} &c_{1} X_{1}\left(t_{0}\right)+c_{2} X_{2}\left(t_{0}\right)=x_{0} \\ &c_{1} \dot{X}_{1}\left(t_{0}\right)+c_{2} \dot{X}_{2}\left(t_{0}\right)=u_{0} \end{aligned} \nonumber \]
which is a system of two linear equations for the two unknowns \(c_{1}\) and \(c_{2}\) . In matrix form,
\[\left(\begin{array}{ll} X_{1}\left(t_{0}\right) & X_{2}\left(t_{0}\right) \\ \dot{X}_{1}\left(t_{0}\right) & \dot{X}_{2}\left(t_{0}\right) \end{array}\right)\left(\begin{array}{l} c_{1} \\ c_{2} \end{array}\right)=\left(\begin{array}{l} x_{0} \\ u_{0} \end{array}\right) . \nonumber \]
We can solve Equation \ref{8.8} for any specified values of \(t_{0}, x_{0}\) and \(u_{0}\) if the 2-by-2 matrix is invertible, and that will be the case if its determinant is nonzero. The determinant is called the Wronskian and is defined by
\[W=X_{1} \dot{X}_{2}-\dot{X}_{1} X_{2} \nonumber \]
In the language of linear algebra, when \(W \neq 0\) the functions \(X_{1}=X_{1}(t)\) and \(X_{2}=X_{2}(t)\) are linearly independent and span the solution space of the secondorder linear differential equation given by Equation \ref{8.4}. The solution space of the ode satisfies the conditions of a vector space and the two solutions \(X_{1}\) and \(X_{2}\) act as basis vectors for this space. The dimension of this space is two, corresponding to the order of the differential equation.
Example: Show that the functions \(X_{1}(t)=\cos \omega t\) and \(X_{2}(t)=\sin \omega t\) have a nonzero Wronskian when \(\omega \neq 0\) .
We have
\[\begin{array}{ll} X_{1}(t)=\cos \omega t, & X_{2}(t)=\sin \omega t \\ \dot{X}_{1}(t)=-\omega \sin \omega t, & \dot{X}_{2}(t)=\omega \cos \omega t \end{array} \nonumber \]
The Wronskian is given by
\[W=\left|\begin{array}{cc} \cos \omega t & \sin \omega t \\ -\omega \sin \omega t & \omega \cos \omega t \end{array}\right|=\omega\left(\cos ^{2} \omega t+\sin ^{2} \omega t\right)=\omega \nonumber \]
so that \(W \neq 0\) when \(\omega \neq 0\) .
Example: Show that the functions \(X_{1}(t)=e^{a t}\) and \(X_{2}(t)=e^{b t}\) have a nonzero Wronskian when \(a \neq b\) .
We have
\[\begin{array}{ll} X_{1}(t)=e^{a t}, & X_{2}(t)=e^{b t} \\ \dot{X}_{1}(t)=a e^{a t}, & \dot{X}_{2}(t)=b e^{b t} \end{array} \nonumber \]
The Wronskian is given by
\[W=\left|\begin{array}{cc} e^{a t} & e^{b t} \\ a e^{a t} & b e^{b t} \end{array}\right|=(b-a) e^{(a+b) t} \nonumber \]
so that \(W \neq 0\) when \(a \neq b\) .
Example: Show that the functions \(X_{1}(t)=1\) and \(X_{2}(t)=t\) have a nonzero Wronskian.
We have
\[\begin{array}{ll} X_{1}(t)=1, & X_{2}(t)=t \\ \dot{X}_{1}(t)=0, & \dot{X}_{2}(t)=1 \end{array} \nonumber \]
The Wronskian is given by
\[W=\left|\begin{array}{ll} 1 & t \\ 0 & 1 \end{array}\right|=1 \nonumber \]