Solving Dy/dx + Y Cot X = 5e^(cos X) With Y(π/2) = -4
Introduction
In this article, we will delve into the process of solving a first-order linear differential equation. Specifically, we will tackle the equation:
\frac{dy}{dx} + y \cot x = 5e^{\cos x}
with the initial condition:
y\left(\frac{\pi}{2}\right) = -4
Differential equations are fundamental tools in various fields, including physics, engineering, economics, and biology, as they describe the relationships between a function and its derivatives. Solving these equations allows us to understand and model dynamic systems. This particular equation is a first-order linear differential equation, a type that can be solved using an integrating factor. We will explore each step in detail, ensuring a clear and comprehensive understanding of the solution process. Understanding differential equations is crucial for anyone working with mathematical models of real-world phenomena. This article aims to provide a step-by-step guide to solving this specific equation, enhancing your understanding of differential equations and their applications. The techniques used here can be applied to a wide range of similar problems, making this a valuable skill for students and professionals alike. This detailed exploration will not only provide the solution but also enhance your grasp of the underlying concepts and methods.
Identifying the Type of Differential Equation
The given differential equation is a first-order linear differential equation. This classification is essential because it dictates the method we use to solve it. A first-order linear differential equation generally takes the form:
\frac{dy}{dx} + P(x)y = Q(x)
where P(x) and Q(x) are functions of x. Comparing this general form to our equation:
\frac{dy}{dx} + y \cot x = 5e^{\cos x}
we can identify that P(x) = cot x and Q(x) = 5e^(cos x). Recognizing this form is the first crucial step in solving the equation. The linear nature of the equation means that the dependent variable y and its derivative appear only to the first power, and there are no products of y and its derivatives. This linearity allows us to use the integrating factor method, which is a systematic approach to finding the solution. The first-order designation indicates that the highest derivative in the equation is the first derivative. Understanding these characteristics is vital because different types of differential equations require different solution techniques. By correctly identifying the equation type, we can choose the most efficient and accurate method to find the solution. This foundational step sets the stage for the subsequent steps in the solution process, ensuring a clear path to the final answer. Therefore, recognizing the equation as a first-order linear differential equation is pivotal for proceeding with the integrating factor method.
Finding the Integrating Factor
To solve the differential equation, we need to find the integrating factor, which we denote as μ(x). The integrating factor is a function that, when multiplied by the differential equation, makes the left-hand side a derivative of a product. This simplifies the equation, allowing us to integrate both sides. The formula for the integrating factor is:
μ(x) = e^{∫ P(x) dx}
In our case, P(x) = cot x. Therefore, we need to compute the integral of cot x with respect to x:
∫ cot x dx = ∫ \frac{\cos x}{\sin x} dx
This integral can be solved using a simple substitution. Let u = sin x, then du = cos x dx. So, the integral becomes:
∫ \frac{1}{u} du = \ln |u| = \ln |\sin x|
Now we can find the integrating factor:
μ(x) = e^{\ln |\sin x|} = |\sin x|
Since we are dealing with the interval around x = π/2, we can assume sin x is positive, so we can take μ(x) = sin x. The integrating factor is a crucial element in solving first-order linear differential equations. By multiplying the entire equation by this factor, we transform the left-hand side into the derivative of a product, making the equation integrable. The choice of sin x as the integrating factor simplifies the subsequent steps and allows us to find a solution more easily. This step highlights the importance of understanding integration techniques and the properties of logarithmic and exponential functions. The correct calculation of the integrating factor is essential for the accuracy of the final solution. By carefully following this procedure, we ensure that the differential equation is correctly transformed and prepared for the next stage of the solution process. This meticulous approach is vital for solving differential equations effectively.
Multiplying the Equation by the Integrating Factor
Now that we have the integrating factor, μ(x) = sin x, we multiply both sides of the original differential equation by it. This step is crucial because it transforms the left-hand side of the equation into the derivative of a product, which we can then easily integrate. The original equation is:
\frac{dy}{dx} + y \cot x = 5e^{\cos x}
Multiplying both sides by sin x, we get:
\sin x \frac{dy}{dx} + y \sin x \cot x = 5e^{\cos x} \sin x
Since cot x = cos x / sin x, the equation simplifies to:
\sin x \frac{dy}{dx} + y \cos x = 5e^{\cos x} \sin x
The left-hand side of this equation is now the derivative of the product y sin x with respect to x. This can be written as:
\frac{d}{dx}(y \sin x) = 5e^{\cos x} \sin x
This step is a significant transformation, as it converts the differential equation into a form that is directly integrable. The multiplication by the integrating factor is the key to this transformation. It cleverly uses the properties of derivatives and trigonometric functions to rewrite the equation in a more manageable form. Recognizing that the left-hand side is a derivative of a product is crucial for the next step, which involves integrating both sides of the equation. This process highlights the power of the integrating factor method in solving first-order linear differential equations. By carefully performing this multiplication and simplification, we set the stage for the final integration, which will yield the general solution to the equation. The clarity and precision in this step are vital for ensuring the accuracy of the solution.
Integrating Both Sides
Having transformed the differential equation into a form where the left-hand side is a derivative, we can now integrate both sides with respect to x. This step allows us to eliminate the derivative and find the general solution. The equation we have is:
\frac{d}{dx}(y \sin x) = 5e^{\cos x} \sin x
Integrating both sides with respect to x, we get:
∫ \frac{d}{dx}(y \sin x) dx = ∫ 5e^{\cos x} \sin x dx
The left-hand side simplifies to y sin x. For the right-hand side, we need to evaluate the integral:
∫ 5e^{\cos x} \sin x dx
This integral can be solved using a substitution. Let u = cos x, then du = -sin x dx. The integral becomes:
∫ -5e^u du = -5e^u + C
Substituting back u = cos x, we have:
-5e^{\cos x} + C
Thus, the equation becomes:
y \sin x = -5e^{\cos x} + C
where C is the constant of integration. The integration step is a fundamental part of solving differential equations. It allows us to move from a derivative form to the actual function. The use of substitution in this case is a common technique for solving integrals involving composite functions. The constant of integration, C, arises because the derivative of a constant is zero, meaning there are infinitely many solutions that differ only by a constant. This constant will be determined by the initial condition given in the problem. This step is crucial for obtaining the general solution, which represents a family of solutions. The accuracy of the integration is paramount, as any error here will propagate through the subsequent steps. Therefore, careful attention to the details of the integration process is essential for arriving at the correct solution.
Applying the Initial Condition
To find the particular solution, we need to apply the initial condition given in the problem. The initial condition is:
y\left(\frac{\pi}{2}\right) = -4
This means that when x = π/2, y = -4. We substitute these values into the general solution we found:
y \sin x = -5e^{\cos x} + C
Substituting x = π/2 and y = -4, we get:
-4 \sin\left(\frac{\pi}{2}\right) = -5e^{\cos(\frac{\pi}{2})} + C
Since sin(π/2) = 1 and cos(π/2) = 0, the equation simplifies to:
-4(1) = -5e^0 + C
-4 = -5(1) + C
-4 = -5 + C
Solving for C, we get:
C = 1
The initial condition is the key to finding a specific solution from the general solution. It provides a fixed point that the solution must pass through, thereby determining the value of the constant of integration. This step is crucial because it narrows down the infinite family of solutions to a single, unique solution that satisfies the given condition. The correct application of the initial condition is essential for solving initial value problems accurately. By substituting the given values and solving for C, we ensure that the solution we obtain is the one that specifically fits the problem's requirements. This process highlights the importance of understanding the relationship between general solutions, particular solutions, and initial conditions in the context of differential equations. Therefore, the careful and precise application of the initial condition is a critical step in finding the final solution.
Writing the Particular Solution
Now that we have found the constant of integration, C = 1, we can write the particular solution to the differential equation. We substitute C = 1 into the general solution:
y \sin x = -5e^{\cos x} + C
y \sin x = -5e^{\cos x} + 1
To solve for y, we divide both sides by sin x:
y = \frac{-5e^{\cos x} + 1}{\sin x}
Thus, the particular solution to the differential equation with the given initial condition is:
y(x) = \frac{1 - 5e^{\cos x}}{\sin x}
This is the final solution to the problem. The particular solution is the unique solution that satisfies both the differential equation and the initial condition. It represents a specific curve in the family of solutions defined by the general solution. The process of finding this solution involves several steps, including identifying the type of equation, finding the integrating factor, multiplying the equation by the integrating factor, integrating both sides, and applying the initial condition. Each step is crucial for the accuracy of the final result. This solution provides a complete and precise answer to the problem, demonstrating the power of the integrating factor method in solving first-order linear differential equations. The final expression for y(x) describes the behavior of the system modeled by the differential equation, providing valuable insights into its dynamics. Therefore, this particular solution is the culmination of the entire solution process and represents the answer to the initial problem.
Conclusion
In this article, we have successfully solved the first-order linear differential equation:
\frac{dy}{dx} + y \cot x = 5e^{\cos x}
with the initial condition:
y\left(\frac{\pi}{2}\right) = -4
We followed a systematic approach, which included identifying the type of differential equation, finding the integrating factor, multiplying the equation by the integrating factor, integrating both sides, and applying the initial condition to find the particular solution. The particular solution we found is:
y(x) = \frac{1 - 5e^{\cos x}}{\sin x}
This process demonstrates the power and effectiveness of the integrating factor method in solving first-order linear differential equations. Understanding and applying these techniques is crucial for solving a wide range of problems in mathematics, physics, engineering, and other fields. The ability to solve differential equations is a fundamental skill for anyone working with mathematical models of real-world phenomena. This article has provided a detailed and comprehensive guide to solving this specific equation, enhancing your understanding of differential equations and their applications. The steps outlined here can be adapted and applied to solve other similar equations, making this a valuable skill for students and professionals alike. By mastering these techniques, you can tackle more complex problems and gain deeper insights into the systems they represent. The journey from the initial equation to the final solution highlights the beauty and utility of mathematical methods in solving real-world problems.
