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(a) Sketch some typical integral curves of the differential equation \(y^{\prime}=y / 2 x.\) (b) Find an equation for the integral curve that passes through the point \((2,1).\)

Short Answer

Expert verified
The solution curve is \( y = \frac{1}{\sqrt{2}}x^{1/2} \), a parabola passing through \((2,1)\).

Step by step solution

01

Understanding the problem

We are given the differential equation \( y' = \frac{y}{2x} \). We need to sketch integral curves and find the particular integral curve passing through the point \((2,1)\).
02

Determine the type of differential equation

The form \( y' = \frac{y}{2x} \) is a separable differential equation since it can be expressed as \( \frac{dy}{dx} = \frac{y}{2x} \). This allows us to separate variables to integrate.
03

Separate the variables

Rewrite the equation as \( \frac{dy}{y} = \frac{1}{2x}dx \). This separates the variables \( y \) and \( x \) on opposite sides, making it ready for integration.
04

Integrate both sides

Integrate both sides: \( \int \frac{dy}{y} = \int \frac{1}{2x}dx \). This results in \( \ln|y| = \frac{1}{2} \ln|x| + C \), where \( C \) is the constant of integration.
05

Simplify the result

Simplify the equation. Using properties of logarithms and exponentiating both sides gives \( |y| = e^C |x|^{1/2} \). Redefine the constant as \(C' = e^C\), leading to \( y = C' x^{1/2} \), considering \( y > 0 \).
06

Find particular solution passing through (2,1)

Substitute \((x, y) = (2, 1)\) into \( y = C’x^{1/2} \) to find \(C’\). Thus, \( 1 = C' \sqrt{2} \), resulting in \( C' = \frac{1}{\sqrt{2}} \). The particular equation is \( y = \frac{1}{\sqrt{2}} x^{1/2} \).
07

Sketch the integral curves

Using the general solution \( y = C' x^{1/2} \), sketch curves for different values of \( C' \) (e.g., \( C' = 1, -1, 0.5, \) etc.). These curves are parabolas opening to the right. Highlight the particular solution \( y = \frac{1}{\sqrt{2}} x^{1/2} \) passing through \((2,1)\).

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Key Concepts

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

Separable Differential Equation
A separable differential equation is a type of ordinary differential equation that can be rearranged so that all terms containing the dependent variable (usually denoted as \(y\)) are on one side, while all terms containing the independent variable (such as \(x\)) are on the other side. This form allows us to solve the equation through integration. The key idea is to "separate" the variables, making it easier to integrate.When faced with a differential equation like \( y' = \frac{y}{2x} \), we recognize it as separable because it can be written as \( \frac{dy}{y} = \frac{1}{2x}dx \). Separating variables is a powerful technique that transforms a differential equation into integrable form. Here are the steps:
  • Identify if the equation can be rearranged into a separable form.
  • Move terms to separate sides of the equation based on their variables.
  • Integrate both sides separately to find the solution.
Separating the equation is a bit like unmixing a cake batter—each ingredient (or variable) has its place, making the process of solving straightforward.
Particular Solution
The particular solution to a differential equation is a unique solution that satisfies not only the differential equation itself but also an initial condition or specific point. In simpler terms, while a general solution contains arbitrary constants, a particular solution provides a specific value for these constants based on given conditions.For instance, in our exercise, a particular solution was sought that passes through the point \((2,1)\). The general solution was found to be \( y = C'x^{1/2} \), but to find the particular solution we substituted \(x = 2\) and \(y = 1\), solving for \(C'\). Here’s more on how it works:
  • Find the general solution first.
  • Use the given point or condition to substitute into the general solution.
  • Calculate the value of the constant to derive the particular solution.
So, the particular solution satisfies not just the underlying relationship itself, but also additional "requirements" like passing through a specific point on its curve.
Constant of Integration
The constant of integration, often denoted as \(C\), appears when solving a differential equation through integration. Since indefinite integration can have any one of infinitely many possible constants, \(C\) represents this family of possible solutions.In the context of our separable differential equation, after integrating, we found an expression involving \( \ln|y| = \frac{1}{2}\ln|x| + C \). This \(C\) stands for the general unknown constant that arises during integration.Here's a deeper look:
  • Every time you integrate, a constant of integration is added because the derivative of a constant is zero—it "could" have been there all along.
  • The constant remains throughout the solution until a particular condition or additional information (an initial condition, for example) allows us to solve specifically for \( C \).
  • Once \(C\) is determined, through initial conditions, it becomes part of the particular solution.
The constant of integration reminds us of the flexibility of mathematical functions—their ability to shift vertically on a graph depending on initial conditions or constraints.

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Most popular questions from this chapter

Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. $$y^{\prime}=\frac{y}{1+y^{2}}$$

A bullet of mass \(m\), fired straight up with an initial velocity of \(v_{0},\) is slowed by the force of gravity and a drag force of air resistance \(k v^{2},\) where \(k\) is a positive constant. As the bullet moves upward, its velocity \(v\) satisfies the equation $$m \frac{d v}{d t}=-\left(k v^{2}+m g\right)$$ where \(g\) is the constant acceleration due to gravity. (a) Show that if \(x=x(t)\) is the height of the bullet above the barrel opening at time \(t,\) then $$m v \frac{d v}{d x}=-\left(k v^{2}+m g\right)$$ (b) Express \(x\) in terms of \(v\) given that \(x=0\) when \(v=v_{0}.\) (c) Assuming that $$v_{0}=988 \mathrm{m} / \mathrm{s}, \quad g=9.8 \mathrm{m} / \mathrm{s}^{2} m=3.56 \times 10^{-3} \mathrm{kg}, \quad k=7.3 \times 10^{-6} \mathrm{kg} / \mathrm{m}$$ use the result in part (b) to find out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.]

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Find a solution to the initial-value problem. \(x^{2} y^{\prime}+2 x y=0, y(1)=2 \quad[\)Hint: Interpret the left-hand side of the equation as the derivative of a product of two functions.]

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