Solving Second Order Differential Equation Showing (x^2+5) D^2y/dx^2 + 4x Dy/dx + 2y = 0
Introduction
In the realm of mathematics, particularly in the study of differential equations, we often encounter intricate relationships between functions and their derivatives. This exploration delves into one such relationship, where we aim to demonstrate a specific equation given a function y defined in terms of x. Specifically, we are given the function $y=\frac{x}{x^2+5}$ and tasked with proving that it satisfies the second-order differential equation $\left(x^2+5\right) \frac{d^2 y}{d x^2}+4 x \frac{d y}{d x}+2 y=0$. This problem beautifully intertwines the concepts of differentiation, algebraic manipulation, and the verification of solutions to differential equations. To embark on this journey, we will systematically compute the first and second derivatives of y with respect to x, and then meticulously substitute these derivatives into the given differential equation. Through careful simplification and algebraic maneuvering, we will ultimately demonstrate the validity of the equation, showcasing the inherent connection between the function and its derivatives. This process not only reinforces our understanding of calculus but also provides a glimpse into the broader applications of differential equations in modeling real-world phenomena.
This article will meticulously guide you through the steps required to solve this problem. We will begin by finding the first derivative, dy/dx, using the quotient rule. Then, we will proceed to calculate the second derivative, d²y/dx², again employing the quotient rule and other differentiation techniques as needed. Finally, we will substitute both derivatives, along with the original function y, into the given equation and simplify to show that the left-hand side equals zero. This exercise not only demonstrates a specific mathematical result but also provides a valuable opportunity to practice and reinforce your skills in differential calculus and algebraic manipulation. By the end of this exploration, you will have a deeper appreciation for the elegance and power of mathematical reasoning in solving complex problems. So, let us embark on this mathematical journey, armed with our knowledge of calculus and a thirst for understanding, and unveil the solution to this intriguing problem.
Step-by-Step Solution: Verifying the Differential Equation
1. Finding the First Derivative (dy/dx)
To begin, we must determine the first derivative of the function $y = \fracx}{x^2 + 5}$ with respect to x. This requires the application of the quotient rule, a fundamental concept in differential calculus. The quotient rule states that if we have a function defined as the quotient of two other functions, say u(x) and v(x), such that $y = \frac{u(x)}{v(x)}$, then the derivative of y with respect to x is given bydx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{[v(x)]^2}$. In our case, we can identify u(x) = x and v(x) = x² + 5. Therefore, we need to find the derivatives of u(x) and v(x) individually. The derivative of u(x) = x with respect to x is simplydx} = 1$. The derivative of v(x) = x² + 5 with respect to x isdx} = 2x$. Now that we have all the necessary components, we can apply the quotient rule to find dy/dx. Substituting the expressions for u(x), v(x), du/dx, and dv/dx into the quotient rule formula, we getdx} = \frac{(x^2 + 5)(1) - (x)(2x)}{(x^2 + 5)^2}$. Next, we simplify the numeratordx} = \frac{x^2 + 5 - 2x2}{(x2 + 5)^2}$. Combining like terms in the numerator, we arrive at the expression for the first derivative{dx} = \frac{5 - x2}{(x2 + 5)^2}$. This result is crucial, as it forms the foundation for calculating the second derivative and ultimately verifying the given differential equation. Understanding and correctly applying the quotient rule is essential for tackling problems of this nature in calculus.
2. Calculating the Second Derivative (d²y/dx²)
Having determined the first derivative, $\fracdy}{dx} = \frac{5 - x2}{(x2 + 5)^2}$, our next task is to compute the second derivative, denoted as $\frac{d2y}{dx2}$. This involves differentiating the first derivative with respect to x. Once again, we encounter a quotient of two functions, making the quotient rule our primary tool. We can consider the numerator as u(x) = 5 - x² and the denominator as v(x) = (x² + 5)². Before applying the quotient rule, we need to find the derivatives of u(x) and v(x). The derivative of u(x) = 5 - x² with respect to x isdx} = -2x$. Finding the derivative of v(x) = (x² + 5)² requires the application of the chain rule. The chain rule states that if we have a composite function, say y = f(g(x)), then its derivative with respect to x is given bydx} = f'(g(x)) \cdot g'(x)$. In our case, we can think of v(x) as a composite function where the outer function is squaring and the inner function is x² + 5. Therefore, applying the chain rule, we getdx} = 2(x^2 + 5)(2x) = 4x(x^2 + 5)$. Now that we have du/dx and dv/dx, we can apply the quotient rule to find the second derivativedx^2} = \frac{(x^2 + 5)^2(-2x) - (5 - x2)(4x(x2 + 5))}{[(x^2 + 5)2]2}$. The next step is to simplify this complex expression. We can factor out (x² + 5) from the numeratordx^2} = \frac{(x^2 + 5)[(x^2 + 5)(-2x) - (5 - x2)(4x)]}{(x2 + 5)^4}$. Now we can cancel one factor of (x² + 5) from the numerator and denominatordx^2} = \frac{(x^2 + 5)(-2x) - (5 - x2)(4x)}{(x2 + 5)^3}$. Next, we expand the terms in the numeratordx^2} = \frac{-2x^3 - 10x - 20x + 4x3}{(x2 + 5)^3}$. Combining like terms in the numerator, we obtain the simplified expression for the second derivative{dx^2} = \frac{2x^3 - 30x}{(x^2 + 5)^3}$. This result, while complex, is a crucial component in verifying the given differential equation. The careful application of both the quotient and chain rules has allowed us to navigate this intricate differentiation process.
3. Substituting into the Differential Equation and Verifying
With the first derivative, $\fracdy}{dx} = \frac{5 - x2}{(x2 + 5)^2}$, and the second derivative, $\frac{d2y}{dx2} = \frac{2x^3 - 30x}{(x^2 + 5)^3}$, in hand, we are now prepared to substitute these expressions, along with the original function $y = \frac{x}{x^2 + 5}$, into the given differential equationd x^2}+4 x \frac{d y}{d x}+2 y=0$. This substitution will allow us to verify whether the function y indeed satisfies the equation. Let's begin by substituting the expressions for y, dy/dx, and d²y/dx² into the left-hand side of the equation(x^2 + 5)^3}\right) + 4x \left(\frac{5 - x2}{(x2 + 5)^2}\right) + 2 \left(\frac{x}{x^2 + 5}\right)$. Our next step is to simplify this complex expression. We can begin by canceling a factor of (x² + 5) in the first term(x^2 + 5)^2} + \frac{4x(5 - x2)}{(x2 + 5)^2} + \frac{2x}{x^2 + 5}$. To combine the terms, we need to find a common denominator. The common denominator is (x² + 5)². Therefore, we need to multiply the last term by (x² + 5)/(x² + 5)(x^2 + 5)^2} + \frac{4x(5 - x2)}{(x2 + 5)^2} + \frac{2x(x^2 + 5)}{(x^2 + 5)^2}$. Now that all terms have the same denominator, we can combine the numerators(x^2 + 5)^2}$. Next, we expand the terms in the numerator(x^2 + 5)^2}$. Now, we combine like terms in the numerator(x^2 + 5)^2}$. This simplifies to(x^2 + 5)^2}$. And finally, we have{x^2 + 5}$ satisfies the differential equation $\left(x^2+5\right) \frac{d^2 y}{d x^2}+4 x \frac{d y}{d x}+2 y=0$. This meticulous process demonstrates the power of calculus and algebraic manipulation in solving differential equations.
Conclusion
In this comprehensive exploration, we have successfully demonstrated that the function $y = \frac{x}{x^2 + 5}$ satisfies the given second-order differential equation $\left(x^2+5\right) \frac{d^2 y}{d x^2}+4 x \frac{d y}{d x}+2 y=0$. This was achieved through a systematic approach involving the application of fundamental calculus concepts, including the quotient rule and the chain rule, for differentiation. We meticulously calculated the first and second derivatives of y with respect to x, and then carefully substituted these expressions, along with the original function, into the differential equation. Through diligent algebraic simplification, we were able to show that the left-hand side of the equation indeed equals zero, thus verifying the solution. This exercise serves as a powerful illustration of how calculus and algebraic techniques can be combined to solve complex mathematical problems. Furthermore, it highlights the importance of understanding and applying differentiation rules accurately, as well as the ability to manipulate algebraic expressions effectively. The process of solving differential equations is not only a crucial aspect of mathematics but also has wide-ranging applications in various fields, such as physics, engineering, and economics, where mathematical models are used to describe and predict real-world phenomena. By mastering these techniques, we equip ourselves with valuable tools for analyzing and understanding the world around us.
Moreover, this exploration reinforces the concept of verifying solutions to differential equations. It is not enough to simply propose a solution; we must rigorously demonstrate that the proposed solution satisfies the equation. This verification process often involves substituting the solution and its derivatives into the equation and showing that the equation holds true. This meticulous approach ensures the validity of our results and provides a solid foundation for further analysis and application. In conclusion, this journey through the solution of a second-order differential equation has not only provided a specific answer but has also reinforced fundamental mathematical principles and problem-solving strategies that are essential for success in mathematics and related fields. The ability to tackle such problems with confidence and precision is a testament to the power of mathematical reasoning and the elegance of calculus.
