To determine whether $y(x) = \frac{6}{5} - e^{-2x}$ is a solution to the differential equation $y' + 20y = 24$, we need to:

1. Calculate the derivative of $y(x)$, i.e., $y'(x)$.
2. Substitute $y(x)$ and $y'(x)$ into the given equation $y' + 20y = 24$ and check if it holds true.

Step 1: Find $y'(x)$

Given that: $y(x) = \frac{6}{5} - e^{-2x}$ Let's differentiate it with respect to $x$: $y'(x) = \frac{d}{dx} \left( \frac{6}{5} - e^{-2x} \right)$ The derivative of $\frac{6}{5}$ is 0 because it's a constant, and the derivative of $-e^{-2x}$ is $2e^{-2x}$ (using the chain rule): $y'(x) = 2e^{-2x}$

Step 2: Substitute into the equation

Now we substitute $y(x) = \frac{6}{5} - e^{-2x}$ and $y'(x) = 2e^{-2x}$ into the differential equation $y' + 20y = 24$.

$y' + 20y = 2e^{-2x} + 20\left( \frac{6}{5} - e^{-2x} \right)$ Simplify the expression: $= 2e^{-2x} + 20 \times \frac{6}{5} - 20e^{-2x}$ $= 2e^{-2x} + 24 - 20e^{-2x}$ $= (2e^{-2x} - 20e^{-2x}) + 24$ $= -18e^{-2x} + 24$

Step 3: Check if the equation holds

For the equation to hold true, we need the expression $-18e^{-2x} + 24$ to equal 24: $-18e^{-2x} + 24 = 24$ Subtract 24 from both sides: $-18e^{-2x} = 0$ This implies that: $e^{-2x} = 0$ However, $e^{-2x}$ is never 0 for any real $x$, so this leads to a contradiction.

Conclusion:

Thus, $y(x) = \frac{6}{5} - e^{-2x}$ is not a solution to the differential equation $y' + 20y = 24$.

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Would you like more details on why this happens or a breakdown of any steps?

Here are some related questions to explore:

1. How would you solve a first-order linear differential equation like $y' + 20y = 24$?
2. What are the general methods to check if a given function is a solution to a differential equation?
3. Can we find an explicit solution to $y' + 20y = 24$ using an integrating factor?
4. How do you handle the case where $e^{ax} = 0$ in differential equations?
5. What are some common types of solutions for linear differential equations?

Tip: When checking if a function is a solution, always differentiate it first and then plug both the function and its derivative back into the original equation to verify.