Problem 10 Obtain the particular solution s... [FREE SOLUTION]

Chapter 2: Problem 10

Obtain the particular solution satisfying the initial condition indicated. $$v(d v / d x)=g ; \text { when } x=x_{0}, v=v_{0}$$

Short Answer

Expert verified

The particular solution is $ v(x) = \sqrt{2g(x - x_0) + v_0^2} $.

Step by step solution

Separate Variables

The given differential equation is $ v \frac{d v}{d x} = g $. This can be rewritten by separating variables: $ v \frac{d v}{d x} = g $. We then separate $ v $ and $ x $: integrate $ v \, dv $ with respect to $ v $ and integrate $ g \, dx $ with respect to $ x $, so $ \int v \, dv = g \int dx $.

Integrate Both Sides

Integrate both sides of the equation: $ \int v \, dv = \int g \, dx $. The integration gives $ (1/2)v^2 = gx + C $, where $ C $ is the constant of integration.

Apply Initial Condition

Use the initial condition $ v = v_0 $ when $ x = x_0 $ to determine $ C $. Substitute into the integrated equation: $ (1/2)v_0^2 = g x_0 + C $. Solve for $ C $: $ C = (1/2)v_0^2 - g x_0 $.

Formulate the Particular Solution

Substitute $ C $ back into the integrated equation: $ (1/2)v^2 = gx + (1/2)v_0^2 - g x_0 $. Rearrange to find $ v(x) $: $ v^2 = 2g(x - x_0) + v_0^2 $.

Solve for Final Expression of v(x)

Take the square root of both sides to solve for $ v(x) $: $ v(x) = \sqrt{2g(x - x_0) + v_0^2} $. This is the particular solution.

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Key Concepts

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

Initial Condition

In differential equations, the "initial condition" is a vital part of finding particular solutions. It provides specific values for the variables, determining the unique solution to a differential equation. Let's break it down:

An initial condition is usually given in the form of an equation indicating the value of the dependent variable for a particular value of the independent variable, such as when $ x = x_0 $, $ v = v_0 $.
It serves as the necessary constraint that changes a general solution, which contains arbitrary constants, into a particular solution, specific to the problem at hand.

Applying the initial condition means substituting these known values back into the integrated equation to solve for the constant of integration. Once this constant is known, it becomes possible to express the particular solution that aligns with the initial scenario set by the problem.

Separation of Variables

"Separation of Variables" is a common method used to solve ordinary differential equations by isolating each variable on a different side of the equation. Here's a simple explanation:

The method is applicable when you can rewrite the differential equation in such a way that all terms involving one variable (say $ v $) are on one side, and all terms involving the other variable (such as $ x $) are on the other.
In our example, the equation $ v \frac{d v}{d x} = g $ can be rearranged by moving all $ v $-related terms to one side and $ x $-related terms to the other side, leading to $ \int v \, dv = g \int dx $.

Through this method, integration on each side can proceed separately, allowing the simplification and eventual integration needed to solve the differential equation. This process builds the foundation towards finding the particular solution by enabling the determination of the constant $ C $ from the initial condition.

Integrating Differential Equations

Integrating differential equations is the step following the separation of variables, and it's essential for transforming a differential equation into a solvable form. Here's what you need to know:

Once the variables are separated, the next step is to integrate both sides of the equation, as demonstrated by $ \int v \, dv = \int g \, dx $.
The integration provides a relationship between the variables, expressed in terms of an equation with an added constant of integration $ C $, such as $ \frac{1}{2}v^2 = gx + C $.

This constant of integration is crucial as it represents the family of solutions to the equation before applying any initial conditions. The role of integration here is to simplify the differential equation into a more manageable and explicit function form that can easily incorporate specific initial conditions to reveal a particular solution.

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91影视

Obtain the particular solution satisfying the initial condition indicated. $$v(d v / d x)=g ; \text { when } x=x_{0}, v=v_{0}$$

Short Answer

Step by step solution

Separate Variables

Integrate Both Sides

Apply Initial Condition

Formulate the Particular Solution

Solve for Final Expression of v(x)

Key Concepts

Initial Condition

Separation of Variables

Integrating Differential Equations

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