For what values of r does the function y = erx satisfy the differential equation y'' + 5y' + 6y = 0?

Answers

Answer 1

Answer:

Answer:

$\huge\boxed{r=-2,-3}$

Step-by-step explanation:

To solve for the values of $r$ where the differentialequation $y'' + 5y' + 6y = 0$ is satisfied by the function $y=e^(rx)$ , we first need to find the first and second derivatives of $y$ with respect to $x$ , treating $r$ as a constant.

$\left[\frac{}{}y\frac{}{}\right]'=\left[\frac{}{}e^(rx)\frac{}{}\right]'$

↓ applying the chain rule to the right side: $\displaystyle \left[\frac{}{}f(x)^a\frac{}{}\right]' = a \cdot f(x)^((a\, -\, 1)) \cdot f'(x)$ where $f(x) = e^x$ and $a = r$

$y'=r\cdot e^((rx \,-\, 1)) \cdot e^x$

↓ simplifying using the exponentbaseproduct rule: $x^a \cdot x^b = x^((a \, +\, b))$

$y' = re^(\left[(rx \,-\, 1)\, +\, 1\right])$

$\boxed{y' = re^(rx)} \ \ \leftarrow \ \ \text{first derivative}$

─────────────────────────────────

↓ taking the derivative of $y$ with respect to $x$

$y'' = \left[\frac{}{} re^(rx)\frac{}{}\right]'$

↓ takingout the constant ( $r$ ) on the right side

$y'' = r\left[\frac{}{} e^(rx)\frac{}{}\right]'$

↓ simplifying by substituting in the first derivative

$y'' = r \cdot y'$

$y'' = r \cdot re^(rx)$

$\boxed{y'' = r^2e^(rx)} \ \ \leftarrow \ \ \text{second derivative}$

Now, we can plug these derivative expressions into the differential equation and solvefor r.

$y'' + 5y' + 6y = 0$

↓ pluggingin the derivativeexpressions (think of $y$ as the zeroth derivative of itself)

$r^2e^(rx) + 5(re^(rx)) + 6(e^(rx)) = 0$

↓ factoringout $e^(rx)$ from the left side

$e^(rx)(r^2 + 5r + 6) = 0$

↓ factoring the second-degree polynomial factor

$e^(rx)(r + 2)(r + 3) = 0$

↓ splitting into 3 equations using the zero product property: $\text{If } ABC = 0,\text{ then } A=0\text{ or }B=0\text{ or }C=0.$

First Equation

$e^(rx)=0$

↓ taking the natural log of both sides

$rx = \ln(0)$

$\implies \text{un}\text{de}\text{fi}\text{ne}\text{d}$

Second Equation

$r+2=0$

↓ subtracting 2 from both sides

$\huge\boxed{r=-2}$

Third Equation

$r+3=0$

↓ subtracting 3 from both sides

$\huge\boxed{r=-3}$

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You spin the spinner twice.9
What is the probability of landing on an 8 and then landing on a 9?
Write your answer as a fraction or whole number.
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Answers

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Step-by-step explanation:

What is equivalent to 2 (b+3c)

Answers

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Find the area under the standard normal curve to the right of $ z=2.33 $. Round your answer to four decimal places, if necessary: Answer

Answers

Answer:

0.0099

Step-by-step explanation:

To find the area under the standard normal curve to the right of $z = 2.33$ , you're essentially looking for the probability that a randomly chosen value from the standard normal distribution is greater than 2.33.

Using a standard normal distribution table or a calculator, you can find this probability directly. The value you're looking for is the complement of the cumulative distribution function (CDF) at $z = 2.33$ , which gives you the area to the left of $z = 2.33$ . To find the area to the right, you subtract this value from 1.

In other words, you want to find $1 - P(Z \leq 2.33)$ , where $Z$ is a standard normal random variable.