We are tasked with verifying that the functions $y = e^{-4x}$ and $y = xe^{-4x}$ are solutions to the differential equation:

$y'' + 8y' + 16y = 0$

Step-by-Step Solution:

We'll check if each of the given functions satisfies this differential equation by calculating their first and second derivatives and substituting them into the equation.

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1. Check if $y = e^{-4x}$ is a solution

Step 1: Compute the first derivative $y'$

The function is: $y = e^{-4x}$

Using the chain rule: $y' = \frac{d}{dx} \left(e^{-4x}\right) = -4e^{-4x}$

Step 2: Compute the second derivative $y''$

Now, differentiate $y' = -4e^{-4x}$: $y'' = \frac{d}{dx} \left(-4e^{-4x}\right) = 16e^{-4x}$

Step 3: Substitute $y, y', y''$ into the differential equation

Now we substitute $y = e^{-4x}$, $y' = -4e^{-4x}$, and $y'' = 16e^{-4x}$ into the equation $y'' + 8y' + 16y = 0$:

$16e^{-4x} + 8(-4e^{-4x}) + 16(e^{-4x}) = 0$ $16e^{-4x} - 32e^{-4x} + 16e^{-4x} = 0$ $(16 - 32 + 16)e^{-4x} = 0$ $0 = 0$

This is true, so $y = e^{-4x}$ is a solution.

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2. Check if $y = xe^{-4x}$ is a solution

Step 1: Compute the first derivative $y'$

The function is: $y = xe^{-4x}$

Apply the product rule: $y' = \frac{d}{dx} \left( x e^{-4x} \right) = \frac{d}{dx}(x) \cdot e^{-4x} + x \cdot \frac{d}{dx}(e^{-4x})$ $y' = e^{-4x} + x(-4e^{-4x}) = e^{-4x} - 4xe^{-4x}$

Step 2: Compute the second derivative $y''$

Now differentiate $y' = e^{-4x} - 4xe^{-4x}$ using the product rule for the second term:

$y'' = \frac{d}{dx}\left( e^{-4x} \right) - \frac{d}{dx}\left( 4xe^{-4x} \right)$ $y'' = -4e^{-4x} - 4\left( e^{-4x} - 4xe^{-4x} \right)$ $y'' = -4e^{-4x} - 4e^{-4x} + 16xe^{-4x}$ $y'' = -8e^{-4x} + 16xe^{-4x}$

Step 3: Substitute $y, y', y''$ into the differential equation

Now we substitute $y = xe^{-4x}$, $y' = e^{-4x} - 4xe^{-4x}$, and $y'' = -8e^{-4x} + 16xe^{-4x}$ into the equation $y'' + 8y' + 16y = 0$:

$(-8e^{-4x} + 16xe^{-4x}) + 8(e^{-4x} - 4xe^{-4x}) + 16xe^{-4x} = 0$ $-8e^{-4x} + 16xe^{-4x} + 8e^{-4x} - 32xe^{-4x} + 16xe^{-4x} = 0$ $( -8 + 8)e^{-4x} + (16 - 32 + 16)xe^{-4x} = 0$ $0e^{-4x} + 0xe^{-4x} = 0$ $0 = 0$

This is true, so $y = xe^{-4x}$ is also a solution.

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Conclusion:

Both $y = e^{-4x}$ and $y = xe^{-4x}$ are solutions to the differential equation $y'' + 8y' + 16y = 0$.

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Would you like more details or have any questions on this process? Here are some related questions you might find interesting:

1. What is the general solution to the differential equation $y'' + 8y' + 16y = 0$?
2. How do you solve a second-order linear homogeneous differential equation with constant coefficients?
3. What role does the characteristic equation play in solving such differential equations?
4. How would you handle a non-homogeneous version of this differential equation?
5. Can these methods be applied to higher-order differential equations?

Tip: When checking if a function is a solution to a differential equation, always compute the derivatives first and substitute them systematically into the equation.