Derivatives And Differential Equations Analysis Of Y=2e^(4x)cos(5x)

This article delves into a comprehensive exploration of derivatives, specifically focusing on a given function and its subsequent derivatives. Our primary objective is to provide a step-by-step analysis of the function y = 2e^(4x)cos(5x), encompassing the calculation of its first and second derivatives. Furthermore, we aim to demonstrate that these derivatives satisfy a particular second-order differential equation. This exploration will not only enhance understanding of differentiation techniques but also illuminate the practical applications of derivatives in solving differential equations.

1. Determining the First Derivative (dy/dx)

In this section, we embark on the crucial task of finding the first derivative (dy/dx) of the given function, y = 2e^(4x)cos(5x). This process involves a careful application of the product rule, a fundamental concept in differential calculus. The product rule states that the derivative of the product of two functions is given by: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Understanding and correctly applying this rule is paramount for accurately computing the first derivative.

To effectively utilize the product rule, we identify two distinct functions within our given equation: u(x) = 2e^(4x) and v(x) = cos(5x). The next step involves finding the individual derivatives of each of these functions. For u(x) = 2e^(4x), we employ the chain rule, which is another essential tool in calculus. The chain rule helps us find the derivative of composite functions, where one function is nested inside another. Applying the chain rule, we find that u'(x) = 8e^(4x). Similarly, for v(x) = cos(5x), we again use the chain rule to obtain v'(x) = -5sin(5x). These individual derivatives are the building blocks for calculating the overall derivative using the product rule.

Now that we have u(x), v(x), u'(x), and v'(x), we can meticulously substitute these values into the product rule formula. This substitution yields the expression for dy/dx. The resulting expression represents the rate of change of the function y with respect to x. This derivative is a new function that provides valuable information about the slope of the original function at any given point. By analyzing the first derivative, we can understand where the original function is increasing, decreasing, or has stationary points. The accuracy of this step is crucial, as any errors here will propagate through the subsequent calculations, affecting the final result and interpretation.

After substituting the values into the product rule formula, we obtain: dy/dx = 8e^(4x)cos(5x) - 10e^(4x)sin(5x). This equation represents the first derivative of the given function. It is a combination of exponential and trigonometric terms, reflecting the nature of the original function. This result is not just a mathematical expression; it is a powerful tool for understanding the behavior of the function y = 2e^(4x)cos(5x). The first derivative allows us to analyze the function's increasing and decreasing intervals, locate critical points, and determine the concavity of the graph. These are fundamental concepts in calculus and have wide-ranging applications in various fields, including physics, engineering, and economics.

2. Calculating the Second Derivative (d²y/dx²)

Having successfully determined the first derivative, our next objective is to compute the second derivative (d²y/dx²). The second derivative provides further insights into the behavior of the original function. Specifically, it helps us understand the concavity of the function's graph, indicating whether the graph is curving upwards or downwards. This information is invaluable in sketching the graph of the function and understanding its overall shape. The process of finding the second derivative involves differentiating the first derivative with respect to x. Since our first derivative is a complex expression involving both exponential and trigonometric functions, this step requires careful application of the product and chain rules.

The first derivative, as we calculated earlier, is given by: dy/dx = 8e^(4x)cos(5x) - 10e^(4x)sin(5x). To find the second derivative, we need to differentiate this expression with respect to x. This involves differentiating each term separately and then combining the results. The first term, 8e^(4x)cos(5x), requires the application of the product rule, similar to what we did in the previous section. We identify two functions within this term: u(x) = 8e^(4x) and v(x) = cos(5x). We then find their individual derivatives: u'(x) = 32e^(4x) and v'(x) = -5sin(5x). Applying the product rule, we get the derivative of the first term.

Similarly, we differentiate the second term, -10e^(4x)sin(5x), using the product rule. Here, we identify u(x) = -10e^(4x) and v(x) = sin(5x). Their derivatives are u'(x) = -40e^(4x) and v'(x) = 5cos(5x). Applying the product rule to this term gives us another component of the second derivative. It is crucial to pay close attention to the signs and coefficients in these calculations to avoid errors. A small mistake in differentiating one term can lead to a significantly different final result. Therefore, meticulousness and attention to detail are paramount in this process.

After differentiating both terms and combining the results, we obtain the second derivative. This derivative is a more complex expression than the first derivative, but it provides deeper insights into the function's behavior. The second derivative tells us about the rate of change of the slope of the original function. A positive second derivative indicates that the function is concave up, while a negative second derivative indicates that the function is concave down. Points where the second derivative is zero are potential inflection points, where the concavity of the function changes. By analyzing the second derivative, we can gain a comprehensive understanding of the function's curvature and shape.

After careful calculation and application of the product and chain rules, the second derivative is found to be: d²y/dx² = 32e^(4x)cos(5x) - 40e^(4x)sin(5x) - 40e^(4x)sin(5x) - 50e^(4x)cos(5x). Simplifying this expression, we get: d²y/dx² = -18e^(4x)cos(5x) - 80e^(4x)sin(5x). This equation represents the second derivative of the given function. It is a critical component in the subsequent steps, where we will verify that it satisfies a given differential equation.

3. Verifying the Differential Equation: d²y/dx² - 8dy/dx + 41y = 0

Having computed both the first and second derivatives, we now arrive at the pivotal task of demonstrating that these derivatives, along with the original function, satisfy a given second-order differential equation. Specifically, we aim to show that d²y/dx² - 8dy/dx + 41y = 0. This process involves substituting the expressions we derived for dy/dx and d²y/dx², as well as the original function y, into the left-hand side of the equation. If the substitution results in the left-hand side simplifying to zero, then we have successfully verified that the function and its derivatives satisfy the differential equation.

This verification process is not just a mathematical exercise; it is a powerful way to confirm the accuracy of our calculations and demonstrate the consistency of the mathematical relationships. Differential equations are fundamental in many areas of science and engineering, as they describe the relationships between a function and its derivatives. Solving differential equations allows us to model and understand a wide range of phenomena, from the motion of objects to the flow of heat. Therefore, the ability to verify solutions to differential equations is a crucial skill in these fields.

We begin by substituting the expressions for y, dy/dx, and d²y/dx² into the left-hand side of the equation. We have: y = 2e^(4x)cos(5x), dy/dx = 8e^(4x)cos(5x) - 10e^(4x)sin(5x), and d²y/dx² = -18e^(4x)cos(5x) - 80e^(4x)sin(5x). Substituting these into the equation d²y/dx² - 8dy/dx + 41y = 0, we get: [-18e^(4x)cos(5x) - 80e^(4x)sin(5x)] - 8[8e^(4x)cos(5x) - 10e^(4x)sin(5x)] + 41[2e^(4x)cos(5x)]. This substitution is a critical step, and it is essential to ensure that each term is correctly placed with the appropriate sign.

Next, we simplify the expression by distributing the constants and combining like terms. This involves multiplying the terms inside the brackets by the constants outside and then grouping the terms with e^(4x)cos(5x) and e^(4x)sin(5x). This simplification process is a meticulous task that requires careful attention to detail. Any errors in this step can lead to an incorrect conclusion about whether the equation is satisfied. The goal is to reduce the expression to its simplest form and see if it equals zero.

After distributing and combining like terms, we obtain: -18e^(4x)cos(5x) - 80e^(4x)sin(5x) - 64e^(4x)cos(5x) + 80e^(4x)sin(5x) + 82e^(4x)cos(5x). Now, we can further simplify by combining the e^(4x)cos(5x) terms and the e^(4x)sin(5x) terms. We have: (-18 - 64 + 82)e^(4x)cos(5x) + (-80 + 80)e^(4x)sin(5x). This simplification reveals a remarkable result.

Finally, we observe that the coefficients of both the e^(4x)cos(5x) and e^(4x)sin(5x) terms sum to zero: (0)e^(4x)cos(5x) + (0)e^(4x)sin(5x) = 0. This confirms that the left-hand side of the equation simplifies to zero, thus verifying that the function y = 2e^(4x)cos(5x) and its derivatives satisfy the differential equation d²y/dx² - 8dy/dx + 41y = 0. This is a significant achievement, as it demonstrates that the function is a solution to the differential equation. This verification process not only validates our calculations but also provides a deeper understanding of the relationship between the function and its derivatives.

Conclusion

In conclusion, this exploration has provided a detailed analysis of the function y = 2e^(4x)cos(5x), encompassing the calculation of its first derivative (dy/dx) and second derivative (d²y/dx²). We successfully applied the product and chain rules to find these derivatives, demonstrating a mastery of fundamental calculus techniques. Furthermore, we rigorously verified that these derivatives, along with the original function, satisfy the second-order differential equation d²y/dx² - 8dy/dx + 41y = 0. This verification process not only confirmed the accuracy of our calculations but also highlighted the importance of differential equations in modeling and understanding various phenomena. This comprehensive analysis underscores the power of calculus in solving complex mathematical problems and its wide-ranging applications in diverse fields.