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Chapter 2: Problem 13

Find the exact solution of the initial value problem. Indicate the interval of existence. $$ y^{\prime}=y / x, y(1)=-2 $$

Short Answer

Expert verified
The solution is \( y = -2x \) with interval of existence \( x \in (0, \infty) \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \( y' = \frac{y}{x} \), which is a first-order linear homogeneous differential equation.
02

Rewrite the equation in separable form

This equation can be rewritten as \( \frac{dy}{dx} = \frac{y}{x} \), which can be separated as \( \frac{dy}{y} = \frac{dx}{x} \).
03

Integrate both sides

Integrate both sides of the equation to get the solution in terms of a constant of integration:\[ \int \frac{dy}{y} = \int \frac{dx}{x} \]This results in:\[ \ln|y| = \ln|x| + C \] where \( C \) is the constant of integration.
04

Solve for y in terms of x

Exponentiate both sides to remove the logarithms:\[ |y| = e^{C} |x| \]Introduce a new constant \( k = \pm e^{C} \), we get:\[ y = kx \].
05

Use initial condition to find the specific solution

Substitute the initial condition \( y(1) = -2 \) into the general solution, \( y = kx \):\[ -2 = k \cdot 1 \]Therefore, \( k = -2 \). The specific solution is \( y = -2x \).
06

Determine the interval of existence

The interval of existence for the solution \( y = -2x \) depends on the domain of the function and the initial value \( x = 1 \). Since the equation involves \( \ln|x| \) during integration, \( x \) cannot be zero. Hence, the interval of existence is \( x \in (0, \infty) \).

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Key Concepts

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

First-Order Linear Differential Equation
A first-order linear differential equation is an equation of the form \( y' + p(x)y = q(x) \), where \( y' \) is the derivative of \( y \) with respect to \( x \), and \( p(x) \) and \( q(x) \) are functions of \( x \). In the context of our exercise, the equation \( y' = \frac{y}{x} \) is first-order because it involves the first derivative of \( y \): \( y' \). It is also linear because the term involving \( y \) (or \( y' \)) is not raised to any power or multiplied with itself. Understanding that this equation is linear helps determine the appropriate method of solving it, often involving the use of integrating factors or separation of variables.
Separable Differential Equations
Separable differential equations are a class of differential equations in which the variables can be separated on opposite sides of the equation. This is exactly the transformation that takes place in our original problem \( y' = \frac{y}{x} \). By rewriting it as \( \frac{dy}{dx} = \frac{y}{x} \), we can separate the variables, resulting in \( \frac{dy}{y} = \frac{dx}{x} \). This conversion allows each side to be independently integrated. It's a common technique, especially powerful due to its simplicity, and often employed when the variables can be naturally split to simplify integration.
Interval of Existence
The interval of existence refers to the range of \( x \) values over which a differential equation's solution remains valid. For our difference equation, we integrate logarithmic functions, which implicate particular domain restrictions. Specifically for \( \ln|x| \), \( x \) cannot be zero, as this would make the function undefined. Given the initial condition \( y(1) = -2 \), we deduced the interval of existence as \( x \in (0, \infty) \). The initial condition also constrains \( x \) to positive values, affirming our interval and ensuring that solutions apply consistently without ambiguity in this region.
Constant of Integration
The constant of integration appears after integrating a differential equation. It arises because integration is an antiderivative operation, which inherently includes an arbitrary constant. In our exercise, after separating variables and integrating both sides, \( \ln|y| = \ln|x| + C \) manifests, where \( C \) is the constant of integration. This factor allows for the flexibility needed to satisfy initial conditions, such that when \( y(1) = -2 \), we can solve for \( C \) through.
  • Substituting back into the equation
  • Introducing a specific solution, such as \( y = -2x \) after determining \( k = \pm e^{C} \)
This determines the precise behavior of the solution, adjusted specifically for the initial condition provided in the problem.

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Most popular questions from this chapter

A tank initially contains \(100 \mathrm{gal}\) of a salt-water solution containing \(0.05 \mathrm{lb}\) of salt for each gallon of water. At time zero, pure water is poured into the tank at a rate of 3 gal per minute. Simultaneously, a drain is opened at the bottom of the tank that allows the salt-water solution to leave the tank at a rate of 2 gal per minute. What will be the salt content in the tank when precisely 50 gal of salt solution remain?

Find the solution of the initial value problem. Discuss the interval of existence and provide a sketch of your solution. $$ y^{\prime}=\cos x-y \sec x, \quad y(0)=1 $$

An autonomous differential equation is given in the form \(y^{\prime}=f(y)\). Perform each of the following tasks without the aid of technology. (i) Sketch a graph of \(f(y)\). (ii) Use the graph of \(f\) to develop a phase line for the autonomous equation. Classify each equilibrium point as either unstable or asymptotically stable. (iii) Sketch the equilibrium solutions in the \(t y\)-plane. These equilibrium solutions divide the ty-plane into regions. Sketch at least one solution trajectory in each of these regions. $$ y^{\prime}=9 y-y^{3} $$

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