Question

Verifying Solutions of Differential Equations Verify that the function \(y=e^{-3 x}+2 x+3\) is a solution to the differential equation \(y^{\prime}+3 y=6 x+11\)

Short answer

Yes, the function satisfies the differential equation.

Step-by-step solution

Differentiate the Function

Start by differentiating the function \(y = e^{-3x} + 2x + 3\) with respect to \(x\). The derivative of \(y\) with respect to \(x\) is found using the sum rule: \[ y' = \frac{d}{dx}(e^{-3x}) + \frac{d}{dx}(2x) + \frac{d}{dx}(3) \]Which gives: \[ y' = -3e^{-3x} + 2 \]

Substitute into the Differential Equation

Substitute \(y\) and \(y'\) into the differential equation \(y' + 3y = 6x + 11\). We have:\[ y' = -3e^{-3x} + 2 \]\[ 3y = 3(e^{-3x} + 2x + 3) = 3e^{-3x} + 6x + 9 \]Substituting these into the equation:\[ (-3e^{-3x} + 2) + (3e^{-3x} + 6x + 9) \]

Simplify and Verify

Simplify the expression:\[ (-3e^{-3x} + 2) + (3e^{-3x} + 6x + 9) = (3e^{-3x} - 3e^{-3x}) + (6x) + (2 + 9) \]\[ = 6x + 11 \]The left-hand side simplifies to \(6x + 11\), which matches the right-hand side of the differential equation \(y' + 3y = 6x + 11\). This confirms that the function is a solution.

Key concepts

Solution Verification

Verifying a solution for a differential equation is like checking if a key fits a lock. You need to confirm that the given function satisfies the condition of the equation. In this case, our original differential equation is \[ y' + 3y = 6x + 11. \]We need to verify the function \( y = e^{-3x} + 2x + 3 \) as a solution. This involves plugging both the function and its derivative back into the equation to see if both sides are equal.
The process involves several key steps:

• Find the derivative \( y' \) of the function.
• Substitute both \( y \) and \( y' \) into the differential equation.
• Simplify both sides of the equation.

This process demonstrates that the function truly fits the equation's requirements, much like finding a perfect match between a key and a lock.

Derivative Calculation

Calculating the derivative is a fundamental step when working with differential equations. In this problem, the function given is\[ y = e^{-3x} + 2x + 3. \]Let's break down the derivative calculation:

• The derivative of \( e^{-3x} \) with respect to \( x \) is calculated using the chain rule, resulting in \( -3e^{-3x} \).
• The derivative of \( 2x \) is straightforward: \( 2 \).
• The derivative of a constant \( 3 \) is simply \( 0 \).

Putting it all together: \[ y' = -3e^{-3x} + 2. \]
Understanding the derivative is crucial as it forms part of the substitution process in verifying the solution of the differential equation. Each component of the expression for \( y \) is differentiated individually, using basic rules of calculus, like the sum and chain rules, to ensure accuracy before moving to the next verification steps.

Substitution in Equations

Once you have both the function and its derivative, the next important step is substitution back into the original differential equation. This step ensures that the left-hand side equals the right-hand side of the equation. With our example function and its derivative already calculated, we substitute these into the equation:
First, write down the differential equation:\[ y' + 3y = 6x + 11. \]You substitute \( y' = -3e^{-3x} + 2 \) and \( 3y = 3(e^{-3x} + 2x + 3) \) which expands to \( 3e^{-3x} + 6x + 9 \):Fill in these expressions into the equation:\[ (-3e^{-3x} + 2) + (3e^{-3x} + 6x + 9) = 6x + 11. \]
Simplification here shows clearly:

• \( -3e^{-3x} + 3e^{-3x} = 0 \)
• \( 2 + 9 = 11 \)

Finally, you get the expression \( 6x + 11 \), which matches the right-hand side of the equation, confirming that the substitution is successful and verifies the solution.