Proving The Differential Equation For Y=2xe^(-3x)

Introduction

In this article, we will delve into a fascinating problem involving differential calculus. Specifically, we aim to demonstrate that if $y = 2xe^{-3x}$, then the second derivative of $y$ with respect to $x$, plus six times the first derivative of $y$ with respect to $x$, plus nine times $y$, equals zero. This problem showcases the application of the product rule and chain rule in differentiation, and it also highlights the concept of a second-order homogeneous differential equation. Understanding such concepts is crucial for various fields, including physics, engineering, and economics, where modeling rates of change is essential. Let's embark on this mathematical journey, breaking down each step to ensure a clear and comprehensive understanding.

Problem Statement

Given the function $y = 2xe^{-3x}$, we are tasked with proving that the following equation holds true:

$$\frac{d^2 y}{dx^2} + \frac{6 dy}{dx} + 9y = 0$$

This problem requires us to find the first and second derivatives of $y$ with respect to $x$, and then substitute these derivatives along with $y$ into the given equation. If the left-hand side of the equation simplifies to zero, we will have successfully demonstrated the validity of the given relationship. This exercise provides a valuable opportunity to reinforce our understanding of differential calculus, particularly the product rule and chain rule, which are fundamental tools for tackling such problems. Solving this problem will not only enhance our mathematical skills but also deepen our appreciation for the elegance and power of calculus in describing and analyzing dynamic systems.

Finding the First Derivative $\frac{dy}{dx}$

To begin, we need to find the first derivative of $y$ with respect to $x$. Given $y = 2xe^{-3x}$, we will apply the product rule, which states that if $y = uv$, then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$. Here, let $u = 2x$ and $v = e^{-3x}$.

The derivative of $u$ with respect to $x$ is:

$$\frac{du}{dx} = \frac{d}{dx}(2x) = 2$$

For the derivative of $v$ with respect to $x$, we apply the chain rule. The chain rule states that if $v = f(g(x))$, then $\frac{dv}{dx} = f'(g(x))g'(x)$. In our case, $v = e^{-3x}$, so we have:

$$\frac{dv}{dx} = \frac{d}{dx}(e^{-3x}) = e^{-3x} \cdot \frac{d}{dx}(-3x) = -3e^{-3x}$$

Now, applying the product rule, we get:

$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = (2x)(-3e^{-3x}) + (e^{-3x})(2) = -6xe^{-3x} + 2e^{-3x}$$

Thus, the first derivative of $y$ with respect to $x$ is $\frac{dy}{dx} = -6xe^{-3x} + 2e^{-3x}$. This step is crucial as it sets the stage for finding the second derivative, which is necessary to verify the given equation. The careful application of the product and chain rules here underscores the importance of mastering these fundamental calculus techniques.

Finding the Second Derivative $\frac{d^2y}{dx^2}$

Next, we need to find the second derivative, $\frac{d^2y}{dx^2}$, which is the derivative of $\frac{dy}{dx}$ with respect to $x$. We found that $\frac{dy}{dx} = -6xe^{-3x} + 2e^{-3x}$. To find the second derivative, we will differentiate this expression again with respect to $x$.

We'll apply the product rule to the first term, $-6xe^{-3x}$, and the chain rule to the second term, $2e^{-3x}$.

For the first term, let $u = -6x$ and $v = e^{-3x}$. Then:

$$\frac{du}{dx} = \frac{d}{dx}(-6x) = -6$$

$$\frac{dv}{dx} = \frac{d}{dx}(e^{-3x}) = -3e^{-3x}$$

Applying the product rule to $-6xe^{-3x}$, we get:

$$\frac{d}{dx}(-6xe^{-3x}) = u\frac{dv}{dx} + v\frac{du}{dx} = (-6x)(-3e^{-3x}) + (e^{-3x})(-6) = 18xe^{-3x} - 6e^{-3x}$$

For the second term, $2e^{-3x}$, we apply the chain rule:

$$\frac{d}{dx}(2e^{-3x}) = 2 \cdot \frac{d}{dx}(e^{-3x}) = 2(-3e^{-3x}) = -6e^{-3x}$$

Now, combining these results, we find the second derivative:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(-6xe^{-3x} + 2e^{-3x}) = (18xe^{-3x} - 6e^{-3x}) + (-6e^{-3x}) = 18xe^{-3x} - 12e^{-3x}$$

Therefore, the second derivative of $y$ with respect to $x$ is $\frac{d^2y}{dx^2} = 18xe^{-3x} - 12e^{-3x}$. This step is critical as it provides us with the final piece of the puzzle needed to verify the original equation. The careful and systematic application of the product and chain rules in this process further solidifies our understanding of these essential calculus techniques.

Verifying the Equation $\frac{d^2 y}{dx^2} + \frac{6 dy}{dx} + 9y = 0$

Now that we have found the first derivative $\frac{dy}{dx} = -6xe^{-3x} + 2e^{-3x}$ and the second derivative $\frac{d^2y}{dx^2} = 18xe^{-3x} - 12e^{-3x}$, we can substitute these expressions, along with the original function $y = 2xe^{-3x}$, into the given equation to verify if it holds true:

$$\frac{d^2 y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$$

Substituting the expressions, we get:

$$(18xe^{-3x} - 12e^{-3x}) + 6(-6xe^{-3x} + 2e^{-3x}) + 9(2xe^{-3x}) = 0$$

Now, we simplify the equation by distributing the constants:

$$18xe^{-3x} - 12e^{-3x} - 36xe^{-3x} + 12e^{-3x} + 18xe^{-3x} = 0$$

Next, we combine like terms:

$$(18xe^{-3x} - 36xe^{-3x} + 18xe^{-3x}) + (-12e^{-3x} + 12e^{-3x}) = 0$$

This simplifies to:

$$0xe^{-3x} + 0e^{-3x} = 0$$

$$0 = 0$$

Thus, the equation is verified. This result demonstrates that the given function $y = 2xe^{-3x}$ satisfies the second-order differential equation $\frac{d^2 y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$. This verification process not only confirms our calculations but also reinforces our understanding of how derivatives and functions relate to each other in the context of differential equations.

Conclusion

In conclusion, we have successfully demonstrated that if $y = 2xe^{-3x}$, then $\frac{d^2 y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$. This problem required us to apply the product rule and chain rule to find the first and second derivatives of the given function. By substituting these derivatives back into the original equation, we were able to verify its validity. This exercise underscores the importance of mastering fundamental calculus techniques and their applications in solving differential equations. Understanding these concepts is crucial for various fields, as differential equations are used to model a wide range of phenomena, from physical systems to economic models. The systematic approach we took, breaking down the problem into manageable steps, is a valuable strategy for tackling complex mathematical challenges. This journey through differentiation and verification not only enhanced our problem-solving skills but also deepened our appreciation for the power and elegance of calculus.