Question

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

Answer 1

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

Explanation:

To solve for the values of
`r` where the differential equation
`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 exponent base product 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]'`

↓ taking out 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 solve for r.

`y'' + 5y' + 6y = 0`

↓ plugging in the derivative expressions (think of
`y` as the zeroth derivative of itself)

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

↓ factoring out
`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}`