Question

From Theorem 2.3.1, the initial value problem $$ y^{\prime}=3 x(y-1)^{1 / 3}, \quad y(0)=9 $$ has a unique solution on an open interval that contains \(x_{0}=0 .\) Find the solution and determine the largest open interval on which it's unique.

Short answer

Question: Determine the unique solution to the given initial value problem (IVP) and the largest open interval containing \(x_0 = 0\) on which the solution is unique. \(\dfrac{dy}{dx} = 3x(y-1)^{1/3}\) with \(y(0) = 9\). Answer: The unique implicit solution of the IVP is \(3(y-1)^{2/3} = \dfrac{3}{2}x^2 + 12\). The largest open interval containing \(x_0 = 0\) on which the solution is unique is \((0, \infty)\).

Step-by-step solution

Identifying the integration method

For the given IVP, we observe that it can be separated into the form of \(\dfrac{dy}{dx} = f(x) g(y)\). This indicates that we can use the method of separation of variables to integrate the equation.

Separating the variables

Rewrite the given IVP as: $$ \dfrac{dy}{dx} = 3x(y-1)^{1/3}. $$ Now separate the variables by placing all terms involving \(y\) on the left and all terms involving \(x\) on the right: $$ \frac{dy}{(y-1)^{1/3}} = 3x dx. $$

Integrating both sides

Integrate both sides of the equation with respect to their respective variables: $$ \int \frac{dy}{(y-1)^{1/3}} = \int 3x dx. $$ The left side integrates to \(3(y-1)^{2/3}\), while the right side integrates to \(\dfrac{3}{2}x^{2}\) plus the constant of integration \(C\): $$ 3(y-1)^{2/3} = \dfrac{3}{2}x^{2} + C. $$

Applying the initial condition

To find the implicit solution, apply the given initial condition \(y(0) = 9\). Plugging the values \(x=0\) and \(y=9\) into the equation above: $$ 3(9-1)^{2/3} = \frac{3}{2}(0)^{2} + C. $$ Solving for \(C\), we get: $$ C = 3(8)^{2/3} = 12. $$

Writing the implicit solution

The implicit solution of the IVP is: $$ 3(y-1)^{2/3} = \dfrac{3}{2}x^2 + 12. $$

Determining the largest open interval

Since the problem asks for the largest open interval containing \(x_0 = 0\), we need to identify any values at which the function is not differentiable. Note that the function is differentiable for all \(y \neq 1\). Thus, the solution is unique on the largest open interval \((0, \infty)\). However, this should be the largest open interval on which it is unique.

Key concepts

Separation of Variables

When solving differential equations, the separation of variables is a widely used method that can simplify many problems into integrable forms. To apply this method, you must have a differential equation that can be written in the form \(\frac{dy}{dx} = f(x)g(y)\). This implies that the derivatives of the function you're looking to find, \(y\), can be expressed as a product of a function of \(x\) and a function of \(y\).

To successfully use separation of variables, you would follow these general steps:

• Rearrange the equation to isolate all terms involving \(y\) (including \(dy\)) on one side, and all terms involving \(x\) (including \(dx\)) on the other side.
• Integrate both sides of the equation, usually resulting in an implicit form of the solution.
• Apply any given initial conditions to solve for the constant of integration, which makes the solution unique.

In the exercise, this approach transforms a potentially complex differential equation into a pair of simpler integrals: one in terms of \(y\) and one in terms of \(x\). Once these integrals are evaluated, the relationship between \(y\) and \(x\) is much clearer.

Implicit Solutions

Solving a differential equation often results in what is called an implicit solution. Unlike an explicit solution that gives \(y\) directly as a function of \(x\), i.e., \(y=f(x)\), an implicit solution involves an equation where \(y\) and \(x\) are intermingled in a way that may not easily or practically be separated.

An implicit form is still a legitimate way of expressing a solution. For differential equations, especially those with initial conditions, it's common to leave the answer in this form due to the complexity that might arise when trying to solve for \(y\) explicitly.

Applying Initial Conditions in Implicit Solutions

One key aspect of an implicit solution is that it allows the incorporation of initial conditions to find a unique solution. For example, given \(y(0) = 9\), you substitute \(x=0\) and \(y=9\) directly into the implicit equation to find the constant of integration \(C\). This incorporation provides a precise solution that satisfies the unique conditions of the problem at hand.

Differential Equations

A differential equation is a mathematical expression that relates a function with its derivatives. In essence, it's an equation that explains how a particular quantity changes with respect to some other variable. They're tremendously important in both mathematics and the sciences because they can model physical phenomena and are used to describe processes such as heat conduction, wave propagation, quantum mechanics, and population dynamics.

Differential equations come in many forms, including ordinary differential equations (ODEs), which relate functions of a single variable to their derivatives, and partial differential equations (PDEs), which involve multiple variables.

Initial Value Problems (IVPs)

An initial value problem (IVP) is a type of differential equation where the solution needs to satisfy a specific initial condition, often a value of the function and its derivatives at a particular point. The main objective when dealing with an IVP is to determine a unique solution that not just satisfies the differential equation itself, but also adheres to the initial conditions provided. This combines the understanding of the equation's behavior with the known starting scenario to arrive at a meaningful solution.