/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Identify the differential equati... [FREE SOLUTION] | 91Ó°ÊÓ

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Identify the differential equation as one that can be solved using only antiderivatives or as one for which separation of variables is required. Then find a general solution for the differential equation. \(\frac{d y}{d x}=\frac{k}{y}\)

Short Answer

Expert verified
The differential equation is separable. The general solution is \( y = \pm \sqrt{2kx + C'} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x} = \frac{k}{y} \). This equation can be identified as a separable differential equation. Such equations can be solved by separating variables, thus allowing us to integrate each side individually.
02

Separate Variables

Rearrange the equation to separate the variables \( y \) and \( x \). Multiply both sides by \( y \) to get \( y \frac{d y}{d x} = k \). Then, multiply both sides by \( dx \) to obtain \( y \, dy = k \, dx \).
03

Integrate Both Sides

Now, integrate both sides of the equation separately. Integrate the left-hand side with respect to \( y \), and the right-hand side with respect to \( x \):\[ \int y \, dy = \int k \, dx \]This gives:\[ \frac{y^2}{2} = kx + C \]where \( C \) is the integration constant.
04

Solve for y (General Solution)

The equation from Step 3 is \( \frac{y^2}{2} = kx + C \). To find the general solution, solve for \( y \):\[ y^2 = 2kx + 2C \]\[ y = \pm \sqrt{2kx + 2C} \]Thus, the general solution is \( y = \pm \sqrt{2kx + C'} \) where \( C' = 2C \) is a constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separable Differential Equations
A separable differential equation is a type of differential equation in which the variables can be separated on different sides of the equation. This means we can rearrange terms so that all expressions involving one variable (for example, \( y \)) are on one side, while all terms involving the other variable (for example, \( x \)) are on the other side. This separation allows each side of the equation to be integrated independently. Separation of variables is a widely used method because it simplifies the integration process. In the given equation \( \frac{d y}{d x} = \frac{k}{y} \), you can identify this as a separable differential equation because you can separate \( y \) and \( x \) as follows: - Multiply both sides by \( y \) to get \( y \frac{d y}{d x} = k \).- Then, multiply both sides by \( dx \) leading to the equation \( y \, dy = k \, dx \). The beauty of separable differential equations lies in their ability to transform complex-looking differential equations into simpler problems which can be solved by basic integration.
Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that reverse differentiation. If a function \( F(x) \) is an antiderivative of a function \( f(x) \), then \( F'(x) = f(x) \). In other words, differentiating the antiderivative will return the original function.In the context of differential equations, finding an antiderivative is crucial because it allows us to integrate terms and eventually solve for the unknown function. For instance, in our separable differential equation example, once we separate variables, we perform integration to find the antiderivative of each side:- The antiderivative of \( y \, dy \) is \( \frac{y^2}{2} + C \), where \( C \) is the constant of integration.- The antiderivative of \( k \, dx \) is \( kx + C' \), with \( C' \) as another constant of integration.Antiderivatives are pivotal in finding the general solution, as they help us reconstruct functions from their derivative forms. This reconstruction is integral to resolving the differential equation.
Integration
Integration is a fundamental concept in calculus, often used to find the area under curves and solve differential equations. In the scope of solving differential equations, integration is used to find antiderivatives and thus progress towards a general solution.In our example, with \( y \, dy = k \, dx \), we integrated both sides separately:- For \( \int y \, dy \), we get \( \frac{y^2}{2} \).- For \( \int k \, dx \), we obtain \( kx + C \).Through integration, we translate the separated differential equation into an equation involving integrals that represent the functions from which our differentials were derived. The process of integration bridges the gap between the components of the differential equation and the general solution.
General Solution
After integration, we usually obtain an equation that includes constants of integration. Solving this integrated equation for the unknown variable gives us the general solution of the differential equation. The general solution offers a family of curves that differentiate into the original differential equation.In our specific case, after integration, we had:\[ \frac{y^2}{2} = kx + C \]To solve for \( y \), we rearrange to find:- \( y^2 = 2kx + 2C \)- \( y = \pm \sqrt{2kx + C'} \), where we set \( C' = 2C \)Thus, the general solution is \( y = \pm \sqrt{2kx + C'} \). This solution encompasses all potential specific solutions for different initial conditions. General solutions are essential as they provide a complete picture of the behavior of differential equations over a range of conditions. Each different \( C' \) represents a particular instance of the problem.

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