12 To find the Wronskian ODE

Since

\[ W\left ( x\right ) =\begin {vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end {vmatrix} =y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }\]

Where \(y_{1},y_{2}\) are two solutions to \(ay^{\prime \prime }+by^{\prime }+cy=0.\) Write

\begin{align*} ay_{1}^{\prime \prime }+py_{1}^{\prime }+cy_{1} & =0\\ ay_{2}^{\prime \prime }+py_{2}^{\prime }+cy_{2} & =0 \end{align*}

Multiply the first ODE above by \(y_{2}\) and the second by \(y_{1}\)

\begin{align*} ay_{2}y_{1}^{\prime \prime }+py_{2}y_{1}^{\prime }+cy_{2}y_{1} & =0\\ ay_{1}y_{2}^{\prime \prime }+py_{1}y_{2}^{\prime }+cy_{1}y_{2} & =0 \end{align*}

Subtract the second from the first

\begin{equation} a\left ( y_{2}y_{1}^{\prime \prime }-y_{1}y_{2}^{\prime \prime }\right ) +p\left ( y_{2}y_{1}^{\prime }-y_{1}y_{2}^{\prime }\right ) =0 \tag {1}\end{equation}

But

\begin{equation} p\left ( y_{2}y_{1}^{\prime }-y_{1}y_{2}^{\prime }\right ) =-pW \tag {2}\end{equation}

And

\begin{align} \frac {dW}{dx} & =\frac {d}{dx}\left ( y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }\right ) \nonumber \\ & =y_{1}^{\prime }y_{2}^{\prime }+y_{1}y_{2}^{\prime \prime }-y_{2}^{\prime }y_{1}^{\prime }-y_{2}y_{1}^{\prime \prime }\nonumber \\ & =y_{1}y_{2}^{\prime \prime }-y_{2}y_{1}^{\prime \prime } \tag {3}\end{align}

Substituting (2,3) into (1) gives the Wronskian differential equation

\begin{align*} -a\left ( \frac {dW}{dx}\right ) -pW & =0\\ aW^{\prime }+pW & =0 \end{align*}

Whose solution is

\[ W\left ( x\right ) =Ce^{-\int \frac {p}{a}dx}\]

Where \(C\) is constant of integration.

Remember: \(W\left ( x_{0}\right ) =0\) does not mean the two functions are linearly dependent. The functions can still be Linearly independent on other interval, It just means \(x_{0}\) can’t be in the domain of the solution for two functions to be solutions. However, if the two functions are linearly dependent, then this implies \(W=0\) everywhere.  So to check if two functions are L.D., need to show that \(W=0\) everywhere.