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Chapter 1: Problem 32

Find a general solution and any singular solutions of the differential equation dymyslashd \(x=y \sqrt{y^{2}-1}\). Determine the points \((a, b)\) in the plane for which the initial value problem \(y^{\prime}=y \sqrt{y^{2}-1}, y(a)=b\) has (a) no solution, (b) a unique solution, (c) infinitely many solutions.

Short Answer

Expert verified
General solution: \( y = \sec(x + C) \). No solution if \( b = \pm 1 \). Unique if \( b \neq \pm 1 \). Infinitely many if on \( y = \pm 1 \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation \( \frac{dy}{dx} = y \sqrt{y^2 - 1} \) is a first-order differential equation. Notice that \( y \sqrt{y^2 - 1} \) suggests it is separable.
02

Separate the Variables

Rewrite the differential equation by separating the variables. We have \( \frac{dy}{y \sqrt{y^2 - 1}} = dx \).
03

Integrate Both Sides

Integrate both sides of the equation. The left side becomes \( \int \frac{1}{y \sqrt{y^2 - 1}} \, dy \) and the right side is \( \int 1 \, dx \).
04

Solve the Integral on the Left

The integral \( \int \frac{1}{y \sqrt{y^2 - 1}} \, dy \) can be solved using a substitution method. Set \( y = \sec(t) \), so \( dy = \sec(t)\tan(t) \, dt \). The integral becomes \( \int \frac{\sec(t)\tan(t)}{\sec(t)\sqrt{\sec^2(t) - 1}} \, dt \) which simplifies to \( \int dt \). Thus, \( t + C = x + C \).
05

Substitute Back

Substituting \( t \) back in terms of \( y \), we get \( \arccos(\frac{1}{y}) = x + C \). Therefore, the general solution is \( y = \sec(x + C) \).
06

Identify Singular Solutions and Critical Points

Singular solutions occur where the denominator of the separated equation is zero, specifically where \( \sqrt{y^2 - 1} = 0 \) or \( y^2 - 1 = 0 \), leading to \( y = \pm 1 \). These are singular solutions and also the points where the initial value problem might have no solution or infinitely many solutions.
07

Analyze Initial Value Problem for Existence and Uniqueness

The initial value problem \( y' = y \sqrt{y^2 - 1} \), with \( y(a) = b \), depends on the relationship of \( b \) to the singular solutions.\( a)\) No solution if \( b = \pm 1 \) and \( (x, y) \) is not on the curve \( y = \pm 1 \). \( b)\) A unique solution if \( y(a) eq \pm 1 \). \( c)\) Infinitely many solutions if the initial condition is exactly on the singular solution \( y = \pm 1 \).

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a type of first-order differential equation. They can be expressed in a form that allows the separation of variables, meaning you can rearrange the equation to isolate the functions of each variable on opposite sides. This makes it easier to solve these types of equations by integrating each side separately. Here's how it works:
  • The general form of a first-order separable differential equation is \( \frac{dy}{dx} = g(x)h(y) \).
  • Rearrange the equation to separate the variables: \( \frac{1}{h(y)} dy = g(x) dx \).
  • This allows us to integrate both sides: \( \int \frac{1}{h(y)} dy = \int g(x) dx \).
  • This results in a solution that often includes an arbitrary constant, usually denoted as \( C \).
In the given problem, \( \frac{dy}{dx} = y \sqrt{y^2 - 1} \), is separable because it can be restructured as \( \frac{dy}{y \sqrt{y^2 - 1}} = dx \). By separating the variables this way, you pave the way to apply integration techniques to find the general solution.
Initial Value Problem
An initial value problem (IVP) is essentially a differential equation coupled with a specific value, or initial condition, for the dependent variable at a particular point. The goal of solving an IVP is to find a particular solution that satisfies both the differential equation and the initial condition.
  • It has the form \( y' = f(x, y) \) with \( y(a) = b \) given.
  • The initial condition \( y(a) = b \) specifies that when \( x = a \), \( y = b \).
  • The solution to the IVP is unique when the function \( f(x, y) \) satisfies certain mathematical conditions, typically related to continuity and smoothness (Lipschitz condition).
  • These conditions ensure that our solution to the differential equation fits the entire context described by the initial condition.
In our problem, we are asked to determine the nature of the solutions \( (a, b) \) based on the relationship of \( b \) with the singular solutions like \( y = \pm 1 \).
Singular Solutions
Singular solutions of a differential equation are specific solutions that do not fit the general solution pattern obtained through separation or other typical methods. These solutions often arise from the structure of the differential equation and represent special cases.
  • A singular solution satisfies the differential equation, but not derived by integrating the separated variables.
  • It can occur at critical points where the mathematical operations required to separate variables become undefined.
  • For example, in our given equation, \( \sqrt{y^2 - 1} = 0 \) when \( y = \pm 1 \). These are singular because the expression within the square root becomes zero, causing the original separation method to fail.
  • These are points where the existence and uniqueness of solutions vary; sometimes causing no solutions or infinitely many solutions.
Understanding singular solutions is crucial in accessing the full dynamics of the differential equations as they reveal points of critical behavior not covered by the general solution.

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Most popular questions from this chapter

The half-life of radioactive cobalt is \(5.27\) years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable? (Ignore the probable presence of other radioactive isotopes.)

Early one morning it began to snow at a constant rate. At 7 A.M. a snowplow set off to clear a road. By 8 A.M. it had traveled 2 miles, but it took two more hours (until \(10 \mathrm{~A}, \mathrm{M}\).) for the snowplow to go an additional 2 miles. (a) Let \(t=0\) when it began to snow and let \(x\) denote the distance traveled by the snowplow at time \(t\). Assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour, say), show that $$ k \frac{d x}{d t}=\frac{1}{t} $$

A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (its half-life is about \(1.28 \times 10^{9}\) years) and that one of every nine potassium atom disintegrations yields an argon atom. What is the age of the rock, measured from the time it contained only potassium?

A river \(100 \mathrm{ft}\) wide is flowing north at \(w\) feet per second. A dog starts at \((100,0)\) and swims at \(v_{0}=4 \mathrm{ft} / \mathrm{s}\), always heading toward a tree at \((0,0)\) on the west bank directly across from the dog's starting point. (a) If \(w=2 \mathrm{ft} / \mathrm{s}\), show that the dog reaches the tree. (b) If \(w=4 \mathrm{ft} / \mathrm{s}\) show that the dog reaches instead the point on the west bank \(50 \mathrm{ft}\) north of the tree. (c) If \(w=6 \mathrm{ft} / \mathrm{s}\), show that the dog never reaches the west bank.

In Problems 17 through 26, first verify that \(y(x)\) satisfies the given differential equation. Then determine a value of the constant \(C\) so that \(y(x)\) satisfies the given initial condition. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation, and highlight the one that satisfies the given initial condition. \(y^{\prime}+3 x^{2} y=0 ; y(x)=C e^{-x^{3}}, y(0)=7\)
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