Chapter 3.4, Problem 66E

To determine

To find: The value of r.

Answer to Problem 66E

The value of $r=2\pm \sqrt{3}$.

Explanation of Solution

Given:

The function $y=e^{rx}$.

The differential equation $y^{″}-4y^{\prime }+y=0$.

Derivative rule:

Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

Calculation:

Obtain the first derivative of $y$.

$\begin{array}{c}{y}^{\prime }\left(x\right)=\frac{d}{dx}\left(e^{rx}\right)\\ =r{e}^{rx}\end{array}$

Therefore, the first derivative of $y$ is $y^{\prime }=r{e}^{rx}$.

Obtain the second derivative of $y=e^{rx}$.

$\begin{array}{c}{y}^{″}\left(x\right)=\frac{d}{dx}\left(y^{\prime }\right)\\ =\frac{d}{dx}\left(r{e}^{rx}\right)\end{array}$

Apply the constant multiple rule (1),

$\begin{array}{c}{y}^{″}\left(x\right)=r\frac{d}{dx}\left(e^{rx}\right)\\ =r\left(r{e}^{rx}\right)\\ =r^2{e}^{rx}\end{array}$

Therefore, the second derivative of $y$ is $\underset{\_}{y^{″}=r^2{e}^{rx}}$.

Substitute the values of $y\left(x\right)$, $y^{\prime }\left(x\right)$ and $y^{″}\left(x\right)$ in $y^{″}-4y^{\prime }+y=0$,

$\begin{array}{l}\left(r^2{e}^{rx}\right)-4\left(r{e}^{rx}\right)+\left(e^{rx}\right)=0\\ e^{rx}\left(r^2-4r+1\right)=0\end{array}$

Since $e^{rx}=0$ does not exit, $r^2-4r+1=0$.

Use the quadratic formula $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with $a=1,b=-4\text{ and }c=1$,

$\begin{array}{c}r=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ =\frac{4\pm \sqrt{16-4}}{2}\\ =\frac{4\pm \sqrt{12}}{2}\\ =\frac{4\pm \sqrt{4\cdot 3}}{2}\end{array}$

On further simplification to obtain the roots of the equation $r^2-4r+1=0$.

$\begin{array}{c}r=\frac{4\pm 2\sqrt{3}}{2}\\ =\frac{2\left(2\pm \sqrt{3}\right)}{2}\\ =2\pm \sqrt{3}\end{array}$

Therefore, the value of $r=2\pm \sqrt{3}$.