Q25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that the m functions   are linearly dependent on (-∞,∞) [Hint: Show thatthese functions are linearly independent if and only if1, x, . . . x^m-1, are linearly independent.] 

Step-by-Step Solution

Answer

The set of functions  $\{e^{rx},x{e}^{rx},x^2{e}^{rx},\mathrm{....},x^{m-1}{e}^{rx}\}$ is linearly dependent on (-∞,∞).

Step 1: Linearly Dependent Functions

Let f(t) and g(t) be differentiable functions. Then they are called linearly dependent if there are nonzero constants c1 and c2 with c1f(t)+c2g(t)=0 for all t. Otherwise they are called linearly independent..

Step 2: Use of Linearly Dependent Functions

We are going to prove by using Linearly Dependent Functions:

 

In order to prove that the set of functions $\{e^{rx},x{e}^{rx},x^2{e}^{rx},\mathrm{....},x^{m-1}{e}^{rx}\}$ is linearly dependent we assume $c_1,c_2,\mathrm{...},c_m$  are constants for which

 

 $c_1{e}^{rx}+c_{2}x{e}^{rx}+c_3{x}^2{e}^{rx}+\mathrm{...}+c_m{x}^{m-1}{e}^{rx}=0$

holds for every x. Knowing that e^rx≠0 stands for every x we can divide the equation with e^rx

Hence,

 $c_1+c_{2}x+c_3{x}^2+\mathrm{...}+c_m{x}^{m-1}=0$

The linear independence of function set $\{1,x,x^2,\mathrm{...},x^{m-1}\}$ was previously proven. Therefore,

 $c_1=c_2=\mathrm{...}=c_m=0$

 

Hence, the final answer is :

 The set of functions  $\{e^{rx},x{e}^{rx},x^2{e}^{rx},\mathrm{....},x^{m-1}{e}^{rx}\}$ is linearly dependent on (-∞,∞).