Question

(a) From Theorem 2.3.1, the initial value problem $$ y^{\prime}=3 x(y-1)^{1 / 3}, \quad y(3)=-7 $$ has a unique solution on some open interval that contains \(x_{0}=3 .\) Determine the largest such open interval, and find the solution on this interval. (b) Find infinitely many solutions of (A), all defined on \((-\infty, \infty)\).

Short answer

The largest open interval for the unique solution is the entire real line, \((- \infty, \infty)\). The solution on this interval is: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 - \frac{3}{2}. $$ (b) How can we find infinitely many solutions of this IVP, all defined on the interval \((- \infty, \infty)\)? There are infinitely many solutions of the form: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 + C, $$ where \(C \neq -\frac{3}{2}\). All these solutions are defined on the interval \((- \infty, \infty)\).

Step-by-step solution

Solving the differential equation

We have the following differential equation: $$ y'(x) = 3x(y-1)^{1/3}. $$ Let's solve this by separating variables: $$ \frac{dy}{(y-1)^{1 / 3}} = 3x dx. $$ Now we can integrate both sides: $$ \int \frac{dy}{(y-1)^{1 / 3}} = \int 3x dx. $$

Integration

Integrating both sides we get: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 + C, $$ where \(C\) is the constant of integration.

Applying the initial condition

Now we apply the initial condition \(y(3) = -7\) to determine \(C\): $$ 3(-7-1)^{2/3} = \frac{3}{2}(3)^2 + C $$ Solve for \(C\): $$ C = 3(-8)^{2/3} - \frac{27}{2} = -\frac{3}{2}. $$ So, the particular solution is: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 - \frac{3}{2}. $$

Finding the largest open interval

To determine the largest open interval where the unique solution exists, we need to find where the function \((y - 1)^{1/3}\) is continuous. Since the cube root function is continuous for all real numbers, the largest open interval is the entire real line, \((-\infty, \infty)\).

(a) Solution and open interval

So, for part (a), the largest open interval is \((-\infty, \infty)\) and the solution on this interval is: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 - \frac{3}{2}. $$

(b) Infinitely many solutions

The given initial condition does not provide enough information to find a unique solution to this problem. However, every other solution can be obtained by changing the value of the constant \(C\). So, there are infinitely many solutions of the form: $$ 3(y-1)^{2/3} = \frac{3}{2}x^2 + C, $$ where \(C \neq -\frac{3}{2}\), and all these solutions are defined on the interval \((-\infty, \infty)\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that describe the relationship between a function and its derivatives. They are a powerful tool in modeling the rates of change in various scientific and engineering fields. For instance, the differential equation in the given exercise, \(y'=3x(y-1)^{1/3}\), represents how the rate of change of the function \(y\) with respect to \(x\) is related to \(x\) itself and the function \(y\).

Solving a differential equation typically involves finding a function or set of functions that satisfy the given equation. Once solved, they can predict future states of a system based on initial conditions. This process often requires techniques such as separation of variables, integrating factors, or more advanced numerical methods.

Separation of Variables

Separation of variables is a common technique for solving ordinary differential equations. It involves rearranging the equation so that each variable and its differential are on opposite sides. For instance, in our problem \(y'=3x(y-1)^{1/3}\), we separate variables to yield \(\frac{dy}{(y-1)^{1/3}} = 3x dx\).

The next step is integrating both sides with respect to their respective variables. This method can only be applied when it is possible to express the differential equation in a form where the variables can be separated, which is particularly useful for first-order differential equations. Understanding this technique is essential for any student tackling differential equations.

Integrating Factor

An integrating factor is a function that, once multiplied by a differential equation, enables it to be rewritten in an integrable form. The goal is to make the left side of the equation the derivative of a product of functions. Although this particular exercise does not require the use of an integrating factor, as separation of variables suffices, having knowledge of integrating factors is invaluable for dealing with non-separable first-order linear differential equations.

Unique Solution

A unique solution of a differential equation is one that satisfies the equation as well as an initial condition. Theorem 2.3.1 referenced in the exercise asserts the existence of such a solution under certain conditions. In our case, the uniqueness of the solution arises from the initial value provided, \(y(3) = -7\).

However, it is essential to understand that not all differential equations have unique solutions; some might have multiple, and others might not have any solutions at all. The existence of a unique solution often relies on certain theorems, which can be based on characteristics like continuity and differentiability of the involved functions.

Initial Condition

Initial conditions in differential equations are specific values provided for the function or its derivatives at a particular point. They are used to find a specific solution from a family of potential solutions. In our problem, the initial condition \(y(3) = -7\) allows us to determine the particular constant of integration \(C\) in the solution.

A well-defined initial value problem will use the given initial conditions to find a unique solution that progresses from that point. Initial conditions are crucial because without them, in many cases, we may not be able to determine the behavior of the system at any other point.