Question

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.

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

Short answer

Solutions are linearly independent.

The differential equation is $y^{\text{'}\text{'}}-2a{y}^{\text{'}}+a^{2}y=0$.

Step-by-step solution

Given information from question

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

Step 2: Definition of Wronskian

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 torole="math" localid="1664299063246" $x$.

Prove that the two solutions are linearly independent

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

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

This means our solution are linearly independent.

Again, the general solutions

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

Therefore, auxiliary equation has only root which is$d=a$therefore, auxiliary equation is

$$\begin{gathered} (D-a)(D-a)y=\left(D^2-2aD+a^2\right)y \\ =0 \end{gathered}$$

and the differential equation itself is $y^{\text{'}\text{'}}-2a{y}^{\text{'}}+a^{2}y=0$.