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Chapter 1: Problem 28

Verify that if \(k\) is a constant, then the function \(y(x) \equiv k x\) satisfies the differential equation \(x y^{\prime}=y\) for all \(x\). Construct a slope field and several of these straight line solution curves. Then determine (in terms of \(a\) and \(b\) ) how many different solutions the initial value problem \(x y^{\prime}=y\), \(y(a)=b\) has -one, none, or infinitely many.

Short Answer

Expert verified
The function satisfies the differential equation, and the slope field shows lines through the origin. For \( y(a)=b \), there is a unique solution if \( a \neq 0 \).

Step by step solution

01

Differentiate the Function

Given the function \( y(x) = kx \), differentiate it with respect to \( x \) to find \( y' \). We have \( y'(x) = \frac{d}{dx}(kx) = k \).
02

Verify Differential Equation

Substitute \( y = kx \) and \( y' = k \) into the differential equation \( xy' = y \). LHS: \( x(k) = kx \). RHS: \( y = kx \). Since LHS = RHS, \( y(x) = kx \) satisfies the differential equation.
03

Analyze Slope Field

The slope field for the equation \( xy' = y \) involves plotting vector slopes at various points \((x, y)\). The slopes are equal to \( \frac{y}{x} \), which results in lines passing through the origin, consistent with \( y = kx \).
04

Graphing Solution Curves

The solution curves for \( y = kx \) are straight lines through the origin with different slopes \( k \). For different values of \( k \), each represents a separate solution passing through different points on the x-y plane.
05

Evaluate Initial Value Problem

For the initial condition \( y(a) = b \), substitute into the solution form \( b = ka \); hence, \( k = \frac{b}{a} \). There is exactly one value of \( k \) for each \( a eq 0 \). Thus, the initial value problem has a unique solution or is undefined for \( a = 0 \).

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Key Concepts

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

Slope Field
A slope field is a visual representation of a differential equation. It is created by plotting small line segments at various points \(x, y\) on a graph, where each segment's slope corresponds to the value of the derivative at that point. For the equation \(xy' = y\), the slope at each point is determined by the ratio \(\frac{y}{x}\).
This results in each line segment having the same direction as a line through the origin. Essentially, the slope field provides a graphical overview of how potential solutions behave across the plane. They help in visualizing the trajectory of various solutions, known as integral curves, without having to solve the equation analytically.
  • Slope field plots are very handy for gaining an intuitive understanding.
  • These plots show where solutions might converge, diverge, or intersect.
  • It's a tool that assists in predicting the behavior of solutions.
Initial Value Problem
An initial value problem is a type of differential equation problem where the solution must satisfy a given initial condition. In our context, it is necessary to find a function \(y(x)\) that not only satisfies the differential equation \(xy' = y\) but also meets the initial condition \(y(a) = b\).
To solve this, we substitute the general solution, \(y = kx\), into the initial condition:
1. \(b = ka\), which gives \(k = \frac{b}{a}\).
2. With these values of \(k\), we can find a specific solution that passes through the point \(a, b\).
The complexity of initial value problems lies in determining the correct parameter (in this case, \(k\)) to satisfy the condition exactly.
  • For every non-zero \(a\), there is exactly one solution.
  • When \(a = 0\), the problem becomes singular and may not have a unique solution.
  • This demonstrates the importance of initial conditions in ensuring the uniqueness of solutions.
Solution Curves
Solution curves in the context of differential equations are graphs of individual solutions. For a differential equation like \(xy' = y\), the solution curves are a family of lines given by \(y = kx\) for different values of \(k\). Each curve represents a different solution.
In simpler terms, if you pick different values of \(k\), you'll get different straight line solutions, all passing through the origin.
  • These curves help us understand the structure of all possible solutions to the differential equation.
  • Depending on the initial condition given, you can pinpoint the exact curve that forms the solution.
  • Solution curves demonstrate that although the solutions are infinite, each initial condition pinpoints a unique solution.
This graphical approach highlights both the diversity and structure within the set of solutions, making it easier to conceptualize complex differential equations.

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Most popular questions from this chapter

Find explicit particular solutions of the initial value problems in Problems 19 through \(28.\) $$ 2 \sqrt{x} \frac{d y}{d x}=\cos ^{2} y, \quad y(4)=\pi / 4 $$

(a) Show that \(y(x)=C x^{4}\) defines a one-parameter family of differentiable solutions of the differential equation \(x y^{\prime}=4 y\) (Fig. 1.1.9). (b) Show that $$ y(x)=\left\\{\begin{aligned} -x^{4} & \text { if } x<0, \\ x^{4} & \text { if } x \geqq 0 \end{aligned}\right. $$ defines a differentiable solution of \(x y^{\prime}=4 y\) for all \(x\), but is not of the form \(y(x)=C x^{4}\). (c) Given any two real numbers \(a\) and \(b\), explain why -in contrast to the situation in part (c) of Problem 47 - there exist infinitely many differentiable solutions of \(x y^{\prime}=4 y\) that all satisfy the condition \(y(a)=b\)

$$ x y^{\prime}=y+\sqrt{x^{2}+y^{2}} $$

Suppose that a falling hailstone with density \(\delta=1\) starts from rest with negligible radius \(r=0 .\) Thereafter its radius is \(r=k t\) ( \(k\) is a constant) as it grows by accretion during its fall. Use Newton's second law- according to which the net force \(F\) acting on a possibly variable mass \(m\) equals the time rate of change \(d p / d t\) of its momentum \(p=m v-\) to set up and solve the initial value problem $$ \frac{d}{d t}(m v)=m g, \quad v(0)=0 $$ where \(m\) is the variable mass of the hailstone, \(v=d y / d t\) is its velocity, and the positive \(y\) -axis points downward. Then show that \(d v / d t=g / 4\). Thus the hailstone falls as though it were under one-fourth the influence of gravity.

Problems deal with a shallow reservoir that has a one-square-kilometer water surface and an average water depth of 2 meters. Initially it is filled with fresh water, but at time \(t=0\) water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month. The well-mixed water in the reservoir flows out at the same rate. Your first task is to find the amount \(x(t)\) of pollutant (in millions of liters) in the reservoir after 1 months. The incoming water has a pollutant concentration of \(c(t)=10\) liters per cubic meter \(\left(\mathrm{L} / \mathrm{m}^{3}\right)\). Verify that the graph of \(x(t)\) resembles the steadily rising curve in Fig. 1.5.9, which approaches asymptotically the graph of the equilibrium solution \(x(t)=20\) that corresponds to the reservoir's long-term pollutant content. How long does it take the pollutant concentration in the reservoir to reach \(10 \mathrm{~L} / \mathrm{m}^{3} ?\)
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