Problem 50 - Advanced Engineering Mathematics

Question

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).

Step-by-Step Solution

Answer

Intervals of definition are \((-1.13232, 1.71299)\) and \((-2.29551, \infty)\).

Step 1: Find General Solution

First, separate the variables and integrate the differential equation \((-2y + y^2) \, dy = (x - x^2) \, dx\). This gives us the solution as \(-y^2 + \frac{1}{3} y^3 = \frac{1}{2} x^2 - \frac{1}{3} x^3 + c\).

Step 2: Implicit Function for Contour Plot

Given the general solution \(f(x, y) = 2y^3 - 6y^2 + 2x^3 - 3x^2\), we can plot the contours by varying the constant \(c\) over a specified range and using a Computer Algebra System (CAS) for visualization.

Step 3: Initial Condition Analysis

For part (b), apply the initial condition \(y(0) = \frac{3}{2}\) and compute \(f\left(0, \frac{3}{2}\right) = -\frac{27}{4}\). This identifies the corresponding contour for \(c\).

Step 4: Find the Interval of Definition

To find the interval of definition, calculate the derivative \(\frac{dy}{dx} = \frac{(x - x^2)}{(y^2 - 2y)}\). The derivative becomes infinite when \(y = 0\) or \(y = 2\). Substitute these into the equation \(2 y^3 - 6 y^2 + 2 x^3 - 3 x^2 = -\frac{27}{4}\) to find critical \(x\) values, determining the interval as \((-1.13232, 1.71299)\).

Step 5: Determine Constant for Second Initial Condition

For part (c), apply the initial condition \(y(0) = -2\) and find \(f(0, -2) = -40\). This contour corresponds to \(c = -40\).

Step 6: Calculate Second Interval of Definition

Using the same derivative \(\frac{dy}{dx}\), determine infinite points for \(y = 0\) and \(y = 2\). By substituting \(y = 0\) in \(2 y^3 - 6 y^2 + 2 x^3 - 3 x^2 = -40\), solve for \(x = -2.29551\), deriving the interval \((-2.29551, \infty)\).

Key Concepts

Separation of Variables

Separation of variables is a method used to solve differential equations. The idea is to rearrange the equation so that each side depends only on one variable. Once separated, both sides of the equation can be integrated independently.

In the given exercise, the differential equation \((-2y + y^2) \, dy = (x - x^2) \, dx\) was separated. By splitting the terms, the equation becomes solvable through integration, leading us to the expression \(-y^2 + \frac{1}{3} y^3 = \frac{1}{2} x^2 - \frac{1}{3} x^3 + c\).

Separation of variables is especially helpful when dealing with ordinary differential equations (ODEs), as it simplifies many complex relationships into integrable parts.

Implicit Differentiation

Implicit differentiation is used when it is difficult or impossible to solve equations explicitly for one variable. Instead of isolating a variable, we differentiate the entire equation with respect to a certain variable, often involving a mix of x and y.

In this problem, implicit differentiation helps find the derivative \(\frac{dy}{dx}\). Starting from the relationship \(2y^3 - 6y^2 + 2x^3 - 3x^2 = c\), we differentiate both sides.
It's crucial since expressions are often intertwined, making traditional differentiation hard. Here, it results in the expression \(y' = \frac{(x-x^2)}{(y^2 - 2y)}\).

This derivative is important for understanding how y changes with respect to x across different parts of the solution curve.

Interval of Definition

The interval of definition for a differential equation solution outlines the region in which the solution is valid or "behaves well." Determining these intervals involves ensuring that no division by zero or other undefined operations occur.

In this context, the intervals are found by analyzing where the denominator of the derivative \(\frac{dy}{dx} = \frac{(x-x^2)}{(y^2 - 2y)}\) becomes zero, leading to infinite slopes or discontinuities.
By setting \(y^2 - 2y = 0\), we find critical y-values, specifically \(y = 0\) and \(y = 2\), which can cause the derivative to become undefined.

Thanks to these constraints, the approximate intervals such as \((-1.13232, 1.71299)\) and \((-2.29551, \infty)\) are identified by solving for x-values, providing valid solution ranges.

Computer Algebra Systems (CAS)

Computer Algebra Systems (CAS) are powerful software tools that help manage and solve complex mathematical equations and plots. They can perform symbolic algebra, calculus, and many other mathematical tasks.

In solving the given differential equations, a CAS can visualize contour plots by varying the constant \(c\), illustrating the relationships between variables effectively.
This graphical insight is vital for understanding the solution's behavior across different parameter values.

Additionally, CAS can solve algebraic equations more accurately than manual computations, allowing for precise determination of intervals of definition and other critical properties.