Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Constructing Differential Equations. Given three functions that ${\mathbf{f}}_1\left(\mathbf{x}\right)\mathbf{,}{\mathbf{f}}_2\left(\mathbf{x}\right),{\mathbf{f}}_3\left(\mathbf{x}\right)$ are each three times differentiable and whose Wronskian is never zero on (a, b), show that the equation 

 

$\left|\begin{array}{cccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)& y\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{=}\mathbf{0}$

 

is a third-order linear differential equation for which $\{{\mathbf{f}}_1\mathbf{,}{\mathbf{f}}_2{\mathbf{f}}_3\}$ is a fundamental solution set. What is the coefficient of y‴ in this equation?

Step-by-Step Solution

Answer

Therefore, the coefficient of y’’’ is,

 $\mathbf{A}\left(\mathbf{x}\right)\mathbf{=}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|$

Step 1: Use the given three functions, f 1 ( x ) ,   f 2 ( x )   f 3 ( x )

Given the equation is,

 $\left|\begin{array}{cccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)& y\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{=}\mathbf{0}\mathrm{.......}\left(\mathbf{1}\right)$

Consider the  $\mathbf{y}\mathbf{=}{\mathbf{f}}_1\left(\mathbf{x}\right)$and Substitute  $\mathbf{y}\mathbf{=}{\mathbf{f}}_1\left(\mathbf{x}\right)$in the above equation to this determinant equation. Observe that 1^st and 4^th column is identical which clearly states $\mathbf{y}\mathbf{=}{\mathbf{f}}_1\left(\mathbf{x}\right)$ is the solution to the differential equation.

 

Similarly, for the functions $\mathbf{y}\mathbf{=}{\mathbf{f}}_1\left(\mathbf{x}\right)\mathbf{,}{\mathbf{f}}_2\left(\mathbf{x}\right)$ and  $\mathbf{y}\mathbf{=}{\mathbf{f}}_1\left(\mathbf{x}\right)\mathbf{,}{\mathbf{f}}_2\left(\mathbf{x}\right)\mathbf{,}{\mathbf{f}}_3\left(\mathbf{x}\right)$are the solution of the differential equation.

 

By assumption these are linearly independent as the Wronskian on the interval [a, b] is zero.

Therefore, $\{{\mathbf{f}}_1\left(\mathbf{x}\right)\mathbf{,}{\mathbf{f}}_2\left(\mathbf{x}\right){\mathbf{f}}_3\left(\mathbf{x}\right)\}$must be a fundamental solution set.

Step 2: Solve for the coefficient of y‴.

Now, the determinate of the equation (1),

 

$\begin{array}{l}\mathbf{y}{(\mathbf{-}\mathbf{1})}^{1+4}\left|\begin{array}{ccc}{\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|\mathbf{+}\mathbf{y}\mathbf{\text{'}}{(\mathbf{-}\mathbf{1})}^{2+4}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|\\ \mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}{(\mathbf{-}\mathbf{1})}^{3+4}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{(\mathbf{-}\mathbf{1})}^{4+4}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|=0\end{array}$

 

Therefore, the coefficient of y’’’ is,

 $\begin{array}{c}\mathbf{A}\left(\mathbf{x}\right)\mathbf{=}{(\mathbf{-}\mathbf{1})}^{4+4}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|\\ \mathbf{=}\left|\begin{array}{ccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\end{array}\right|\end{array}$

 

Hence Proved.