Question

For what values of r does the function y = erx satisfy the differential equation y'' − 6y' + 2y = 0? (enter your answers as a comma-separated list.)

Answer 1

The values of $r$ is $\boxed{r = 3 - \sqrt7 }$ and $\boxed{r = 3 + \sqrt 7 }.$

Further explanation:

Given:

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

The differential equation is $y'' - 6y' + 2y = 0\,\,\,\,\,or\,\,\,\,\dfrac{{{d^2}y}}{{d{x^2}}} - 6\dfrac{{dy}}{{dx}} + 2y = 0$

Explanation:

The given function can be expressed as follows,

$y = {e^{rx}}$

Differentiate the above equation with respect to $x$.

$\begin{aligned}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{rx}}} \right)\\&= {e^{rx}} \times \frac{d}{{dx}}\left( {rx} \right)\\&= {e^{rx}} \times r\\&= r{e^{rx}}\\\end{aligned}$

Again differentiate with respect to $x$.

$\begin{aligned}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {r{e^{rx}}} \right)\\&= r\frac{d}{{dx}}\left( {{e^{rx}}} \right)\\&=r\times {e^{rx}} \times\frac{d}{{dx}}\left( {rx} \right)\\&= {r^2}{e^{rx}}\\\end{aligned}$

Now solve the differential equation.

$\begin{aligned}y'' - 6y' + 2y &= 0\\{r^2}{e^{rx}} - 6r{e^{rx}} + 2{e^{rx}} &= 0\\{e^{rx}}\left( {{r^2} - 6r + 2} \right) &= 0\\{r^2} - 6r + 2 &= 0\\\end{aligned}$

Solve the quadratic equation ${r^2} - 6r + 2 = 0.$

$\begin{aligned}D&= {b^2} - 4ac\\&={\left( { - 6} \right)^2} - 4\left( 1 \right)\left( 2 \right)\\&= 36 - 8\\&= 28\\\end{aligned}$

The value of $r$ can be obtained as follows,

$\begin{aligned}r&= \frac{{ - b \pm \sqrt D }}{{2a}}\\r&=\frac{{ - \left( { - 6} \right) \pm \sqrt {28} }}{{2 \times 1}} \\ r &= \frac{{6 \pm 2\sqrt7 }}{2}\\r&= 3 \pm \sqrt7\\r&= 3 - \sqrt 7 \,\,\,\,\,\,\,or\,\,\,\,\,\,\,r = 3 + \sqrt7 \\\end{aligned}$

The values of $r$ is $\boxed{r = 3 - \sqrt7 }$ and $\boxed{r = 3 + \sqrt 7 }.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Derivatives

Keywords: Derivative, value of x, function, differentiate, minimum value, dy, compute, given value of $x, y=6x, x=4, x=1$

Answer 2

You mean y = e^(rx)? Yes, there end up being two values. They should both work. Nobody said the solution had to be unique! 

Use this expression for y to calculated y', then take the derivative again to get y ' ' . 
y' = r e^(rx) y' ' = r^2 e^(rx) 

So you have: r^2 e^(rx) - 6r e^(rx) + 2e^(rx) = 0 
Since e^(rx) can never be 0, we can divide both sides by e^(rx) to get 

r^2 - 6r + 2 = 0  
r1 =(6-√28)/2=3-√ 7 = 0.354 
r2 =(6+√28)/2=3+√ 7 = 5.646

so r = 0.354, 5.646