Q32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Let y₁x₂= Ce^rx, where C (≠0) and r are real numbers,be a solution to a differential equation. Supposewe cannot determine r exactly but can only approximateit by $r$ . Let   (x) =Ce^rxand consider the error

 $\left|y\left(x\right)-\tilde{y}\left(x\right)\right|$

(a) If r and  $\tilde{r}$ are positive, r ≠­ , show that the errorgrows exponentially large as x approaches + ∞.

(b) If r and  $r$ are negative, r≠ , show that the errorgoes to zero exponentially as x approaches + ∞.

Step-by-Step Solution

Answer

E(x) goes to zero exponentially as  $x\to +\infty$ .

Step 1: Differential Equation

A differential equation is an equation that relates one or more unknown functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two.

Step 2: Proving that E(x) goes to zero exponentially as x → + ∞

Let 

 $y\left(x\right):=C{e}^{rx}$

be a solution of some D.E

such that C≠0 and r ∈ R

 

We to need to find value of r by approximation method.

Let us consider y₁ which is defined as,

 $y_1\left(x\right):=C{e}^{r_{1}x}$

and

 $E\left(x\right)=\left|y\left(x\right)-y_1\left(x\right)\right|$

be a error term.

 

(a): If r>0, r₁>0 and r≠r₁

To Show:  $E\left(x\right)\to +\infty$ as  $x\to +\infty$ Consider,

 $\begin{array}{l}E\left(x\right)=\left|y\left(x\right)-y_1\left(x\right)\right|\\ =|C{e}^{rx}-C{e}^{r_{1}x}|\\ =\left|C(e^{rx}-e^{r_{1}x})\right|\\ =\left|C\right||e^{rx}-e^{r_{1}x}|\\ \ge C\left|e^{rx}\right|-\left|e^{r_{1}x}\right|\end{array}$

 

We know, $e^x\to +\infty$   as  $x\to +\infty$

Therefore $\left|e^{rx}\right|\to +\infty$  as  $x\to +\infty$  and $\left|e^{r_{1}x}\right|\to +\infty$ as  $x\to +\infty$

Hence,  $E\left(x\right)\to +\infty$  as  $x\to +\infty$

(b) If r<0, r₁<0 and r≠r₁

 To Show:$E\left(x\right)\to +\infty$   as  $x\to +\infty$

Consider,

 $\begin{array}{l}E\left(x\right)=\left|y\left(x\right)-y_1\left(x\right)\right|\\ =|C{e}^{rx}-C{e}^{r_{1}x}|\\ =\left|C(e^{rx}-e^{r_{1}x})\right|\\ =\left|C\right||e^{rx}-e^{r_{1}x}|\\ \ge C\left|e^{rx}\right|-\left|e^{r_{1}x}\right|\end{array}$

Since r and r₁ are negative 

 

 $\begin{array}{l}E\left(x\right)=\left|C\right||e^{-rx}-e^{-r_{1}x}|\\ \ge \left|C\right|(\left|e^{-rx}\right|-\left|e^{-r_{1}x}\right|)\\ =C[\frac{1}{\left|e^{-rx}\right|}-\frac{1}{\left|e^{-r_{1}x}\right|}]\end{array}$

Therefore,  $E\left(x\right)\to +\infty$  as  $x\to +\infty$

 

Hence, E(x) goes to zero exponentially as  $x\to +\infty$ .