Recall from Chapter 3, equation (8.5), that a set of functions is linearly independent if their Wronskian is not identically zero. Calculate the Wronskian of each of the following sets to show that in each case they are linearly independent. For each set, write the differential equation of which they are solutions. Also note that each set of functions is a set of basis functions for a linear vector space (see Chapter 3, Section 14, Example 2) and that the general solution of the differential equation gives all vectors of the vector space.

$e^{ax},x{e}^{ax},x^2{e}^{ax}$

Given equation is $e^{ax},x{e}^{ax},x^2{e}^{ax}$.

Definition of Wronskian:

A mathematical determinant whose first row consists of$n$functions of$x$and whose following rows consist of the successive derivatives of these same functions with respect to$x$.

To prove that the two solutions are linearly independent find the Wronskian, and if it was no identically zero, that they are independent

$W=\left|\begin{array}{ccc}{e}^{ax}& x{e}^{ax}& x^2{e}^{ax}\\ a{e}^{ax}& e^{ax}+ax{e}^{ax}& 2x{e}^{ax}+a{x}^2{e}^{ax}\\ a^2{e}^{ax}& a{e}^{ax}+a\left(e^{ax}+ax{e}^{ax}\right)& 2\left(e^{ax}+ax{e}^{ax}\right)+a\left(2x{e}^{ax}+a{x}^2{e}^{ax}\right)\end{array}\right|$

which means the solutions are linearly independent. Therefore, the general solution is

$$\begin{gathered} y=e^{ax}+x{e}^{ax}+x^2{e}^{ax} \\ =\left(1+x+x^2\right)e^{ax} \end{gathered}$$

Now, since it has three independent solution, this means that the solution for third order differential equation. But first left $f=1+x+x^2$, therefore,

$$\begin{gathered} y^{\text{'}}=f^{\text{'}}{e}^{ax}+ay \\ {y}^{\text{'}\text{'}}=f^{\text{'}\text{'}}{e}^{ax}+a{f}^{\text{'}}{e}^{ax}+a{y}^{\text{'}} \\ =f^{\text{'}\text{'}}{e}^{ax}+a\left(y^{\text{'}}-ay\right)+a{y}^{\text{'}} \end{gathered}$$

Solve further,

$$\begin{gathered} =f^{\text{'}\text{'}}{e}^{ax}+2a{y}^{\text{'}}-ay \\ {y}^{\text{'}\text{'}\text{'}}=f^{\text{'}\text{'}\text{'}}{e}^{ax}+a{f}^{\text{'}\text{'}}{e}^{ax}+2a{y}^{\text{'}}-a{y}^{\text{'}} \\ =f^{\text{'}\text{'}\text{'}}{e}^{ax}+a\left(y^{\text{'}\text{'}}-2a{y}^{\text{'}}+ay\right)+2a{y}^{\text{'}\text{'}}-a{y}^{\text{'}} \\ =f^{\text{'}\text{'}}{e}^{ax}+3a{y}^{\text{'}\text{'}}-y^{\text{'}}\left(2a^2+a\right)+ay \end{gathered}$$

Here$f^{\text{'}}=1+2x,f^{\text{'}\text{'}}=2,f^{\text{'}\text{'}\text{'}}=0$.

Therefore, differential equation is $y^{\text{'}\text{'}\text{'}}=3a{y}^{\text{'}\text{'}}+y^{\text{'}}\left(2a^2+a\right)+ay$.