Problem 39 - Advanced Engineering Mathematics

Question

(a) All solutions of the form \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) satisfy the initial condition. In this case, since \(4 / x\) is discontinuous at \(x=0,\) the hypotheses of Theorem 1.1 are not satisfied and the initial-value problem does not have a unique solution. (b) The differential equation has no solution satisfying \(y(0)=y_{0}, y_{0}>0\) (c) In this case, since \(x_{0}>0,\) Theorem 1.1 applies and the initial-value problem has a unique solution given by \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) where \(c=y_{0} / x_{0}^{4}-x_{0} e^{x_{0}}+e^{x_{0}}\)

Step-by-Step Solution

Answer

The equation needs \( x_0 > 0 \) for a unique solution, no unique solution if \( x = 0 \).

Step 1: Understanding the equation

The given differential equation is a form that involves functions of polynomials and exponential functions. It is written as \( y = x^5 e^x - x^4 e^x + c x^4 \). We need to analyze when this equation can satisfy initial conditions uniquely.

Step 2: Analyzing discontinuity at x = 0

Since the term \( \frac{4}{x} \) is discontinuous at \( x = 0 \), the equation does not satisfy the continuity condition. This means any initial-value problem with \( x = 0 \) does not guarantee a unique solution, as depicted in part (a).

Step 3: Checking for initial condition y(0)=y0, y0>0

In part (b), the problem states that there is no solution satisfying \( y(0) = y_0, \) \( y_0 > 0 \). This is because at \( x = 0 \), the function loses certain properties (due to the discontinuity) necessary for the solution to exist.

Step 4: Applying Theorem 1.1 for x0 > 0

For part (c), where \( x_0 > 0 \), Theorem 1.1 guarantees that the differential equation can have a unique solution. The solution is given by the equation \( y = x^5 e^x - x^4 e^x + c x^4 \), where the constant \( c \) is determined by a given point \( (x_0, y_0) \). Here, \( c = \frac{y_0}{x_0^4} - x_0 e^{x_0} + e^{x_0} \), compensating for the initial conditions.

Key Concepts

Initial-Value Problem

An initial-value problem in differential equations involves finding a function that not only satisfies the differential equation but also meets a specific initial condition. This condition typically specifies the value of the unknown function at a particular point.
For instance, if we have a differential equation given by the solution form \( y = x^5 e^x - x^4 e^x + c x^4 \), our task can be finding a function such that \( y(x_0) = y_0 \) for specified \( x_0 \) and \( y_0 \).
This means we're not just looking for a general solution to the differential equation, but a specific one that passes through the point \((x_0, y_0)\).

• This is crucial because the initial condition helps in determining a unique solution, provided that certain conditions are met.
• The uniqueness can sometimes be affected by factors like the continuity and differentiability of the coefficients of the differential equation.

Discontinuity

Discontinuity occurs in a function when there is a sudden jump or break in the graph, meaning the function is not seamless and smooth. In mathematics, a discontinuous point is where a function is not defined or behaves erratically.
For example, in the equation involved in our problem, the presence of the term \( \frac{4}{x} \) leads to a discontinuity at \( x = 0 \).

• Because the function \( \frac{4}{x} \) is undefined at \( x = 0 \), the conditions needed for the solutions to be continuous (and therefore ideally) become violated.
• This makes it impossible to guarantee a unique solution for initial conditions with \( x = 0 \).

The concept of discontinuity is also crucial because it can limit the applicability of certain theorems, such as Theorem 1.1, which might require continuous functions over their domains to assert the uniqueness of solutions.

Unique Solution

A unique solution exists for an initial-value problem when only one function meets both the differential equation and the initial condition.
In the context of this exercise, if \( x_0 > 0 \) and the initial condition \( y(x_0) = y_0 \) are applicable, Theorem 1.1 assures us that the solution \( y = x^5 e^x - x^4 e^x + c x^4 \) is unique.
This is because, away from discontinuities, conditions like continuous derivatives are met, allowing such theorems to hold.

• The constant \( c \) is specifically tailored to the initial condition, calculated using the equation \( c = \frac{y_0}{x_0^4} - x_0 e^{x_0} + e^{x_0} \).
• This makes sure that the curve not only solves the differential equation but also passes precisely through the specified initial point.

Unique solutions are critical as they ensure the reliability and predictability of the mathematical model, ensuring real-world applications can depend solely on one specific outcome.