Problem 40 - Advanced Engineering Mathematics

Question

(a) Writing the system as \(x^{\prime}=x\left(x^{3}-2 y^{3}\right)\) and \(y^{\prime}=y\left(2 x^{3}-y^{3}\right)\) we see that (0,0) is a critical point. Setting \(x^{3}-2 y^{3}=0\) we have \(x^{3}=2 y^{3}\) and \(2 x^{3}-y^{3}=4 y^{3}-y^{3}=3 y^{3} .\) Thus, (0,0) is the only critical point of the system. (b) From the system we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{2 x^{3} y-y^{4}}{x^{4}-2 x y^{3}}$$ or $$\left(2 x^{3} y-y^{4}\right) d x+\left(2 x y^{3}-x^{4}\right) d y=0$$ which is homogeneous. If we let \(y=u x\) it follows that $$\begin{aligned} \left(2 x^{4} u-x^{4} u^{4}\right) d x+\left(2 x^{4} u^{3}-x^{4}\right)(u d x+x d u) &=0 \\ x^{4} u\left(1+u^{3}\right) d x+x^{5}\left(2 u^{3}-1\right) d u &=0 \\ \frac{1}{x} d x+\frac{2 u^{3}-1}{u\left(u^{3}+1\right)} d u &=0 \\ \frac{1}{x} d x+\left(\frac{1}{u+1}-\frac{1}{u}+\frac{2 u-1}{u^{2}-u+1}\right) d u &=0 \end{aligned}$$ Integrating gives $$\ln |x|+\ln |u+1|-\ln |u|+\ln \left|u^{2}-u+1\right|=c_{1}$$ or $$\begin{aligned} x\left(\frac{u+1}{u}\right)\left(u^{2}-u+1\right) &=c_{2} \\ x\left(\frac{y+x}{y}\right)\left(\frac{y^{2}}{x^{2}}-\frac{y}{x}+1\right) &=c_{2} \\ \left(x y+x^{2}\right)\left(y^{2}-x y+x^{2}\right) &=c_{2} x^{2} y \\ x y^{3}+x^{4} &=c_{2} x^{2} y \\ x^{3}+y^{2} &=3 c_{3} x y \end{aligned}$$ (c) We see from the graph that (0,0) is unstable. It is not possible to classify the critical point as a node, saddle, center, or spiral point.

Step-by-Step Solution

Answer

The critical point is (0,0), and it is unstable; no further classification is possible.

Step 1: Identifying Critical Points

To find the critical points of the system given by \( x' = x(x^3 - 2y^3) \) and \( y' = y(2x^3 - y^3) \), we set \( x' = 0 \) and \( y' = 0 \). This implies \( x(x^3 - 2y^3) = 0 \) and \( y(2x^3 - y^3) = 0 \). Since both terms involve either \( x \) or \( y \), we can quickly find that \( (0,0) \) is a critical point.

Step 2: Solving the Homogeneous Differential Equation

We are given the differential equation in the form \( \frac{dy}{dx} = \frac{2x^3y - y^4}{x^4 - 2xy^3} \), which can be rewritten as \((2x^3y - y^4)dx + (2xy^3 - x^4)dy = 0\). By substituting \( y = ux \), we transform this into a homogeneous equation.

Step 3: Applying the Substitution

Using \( y = ux \), replace \( dy = u dx + x du \) in the differential equation leading to terms like \((2x^4u - x^4u^4)dx + (2x^4u^3 - x^4)(udx + xdu) = 0\). After simplifying, we factor out leading terms and organize the equation as \( \frac{1}{x}dx + \frac{2u^3 - 1}{u(u^3 + 1)}du = 0 \).

Step 4: Integrating Both Sides

Integrate both sides to find the implicit solution. For \( \frac{1}{x}dx \), the integral is \( \ln|x| \). The second part, \( \int \left( \frac{1}{u+1} - \frac{1}{u} + \frac{2u-1}{u^2-u+1} \right)du \), is integrated separately, giving \( \ln|u+1| - \ln|u| + \ln|u^2-u+1| \). Combine them to form the equation \( \ln|x| + \ln|u+1| - \ln|u| + \ln|u^2-u+1| = c_1 \).

Step 5: Rewriting the Solution

Transform the logarithmic expression back into exponential form and simplify: \( x\frac{u+1}{u}(u^2-u+1) = c_2 \). Substitute back \( u = \frac{y}{x} \) to express the solution in terms of \( x \) and \( y \), leading to \( (xy + x^2)(y^2 - xy + x^2) = c_2 x^2y \). This rearranges to \( x^3 + y^2 = 3c_3xy \).

Step 6: Analyzing Stability

By observing the behavior of the system near critical points using the graph or phase plane, determine that (0,0) is unstable. However, classification into node, saddle, center, or spiral is not possible with given information.

Key Concepts

Homogeneous Differential Equation

In the realm of differential equations, a homogeneous differential equation is one where each term is a function of the dependent and independent variables of the same degree. That means when you divide or multiply all the terms by a certain factor, the degree of each term remains the same. Here, the form provided is

• \( \frac{dy}{dx} = \frac{2x^3y - y^4}{x^4 - 2xy^3} \)

This equation is homogeneous because both the numerator and the denominator are homogeneous functions of the same degree.
By performing the substitution \( y = ux \), where \( u \) is a function of \( x \), the equation transforms into a separable form, which simplifies the process of solving it. This kind of substitution is common when dealing with homogeneous differential equations, as it exploits the symmetry present in these problems.
The next step after substitution is simplifying the equations, sometimes involving factoring or re-arranging terms to isolate \( du \) and \( dx \), making the equation easier to integrate.

Critical Points

Critical points are solutions to the differential equation where the derivative equals zero. They play a crucial role in understanding the behavior of the system.
In the exercise, the critical point identified is \((0,0)\). This point indicates where the rate of change for both \( x' \) and \( y' \) is zero, suggesting a point of equilibrium.
To find critical points analytically, you set the derivatives \( x' = 0 \) and \( y' = 0 \). Then, you solve the resulting algebraic equations:

• From \( x' = x(x^3 - 2y^3) \), \((x = 0)\) or \((x^3 - 2y^3 = 0)\)
• From \( y' = y(2x^3 - y^3) \), \((y = 0)\) or \((2x^3 - y^3 = 0)\)

The solutions to these equations are the critical points. For this system, solving them verifies that \((0,0)\) is the critical point.

Stability Analysis

Stability analysis helps us understand how a system behaves near its critical points. We examine whether small disturbances or deviations return to the critical point or move away from it.
In our system, it's essential to assess the stability of the critical point \((0,0)\). We observe its behavior by analyzing solutions near this point.
Stability can be determined through different techniques, such as linearization or using phase planes. However, in this case, the analysis suggests that the critical point \((0,0)\) is unstable. Unstable points often mean that any small change will result in significant deviations from the equilibrium state over time.

• As demonstrated, classification into specific types of critical points, like nodes or saddles, wasn't possible, indicating complexity in the system's dynamics.

Phase Plane Analysis

The phase plane offers a graphical way to study differential equations, particularly systems of two equations like the one in the exercise. It provides a visual representation of the trajectories of the system as it evolves over time.
In the phase plane, each point represents a state of the system, and a trajectory shows how these states evolve. Analyzing these trajectories helps identify stable and unstable regions, as well as periodic behaviors such as cycles.
For the given system, the phase plane can indicate the nature of the critical point \((0,0)\). Although the exercise mentions that it is unstable, the phase plane allows us to visualize how solutions spread out from this point. This analysis helps in identifying regions with similar behaviors and understanding the overall dynamics of the differential system.

• The phase plane is invaluable for providing insights that aren't always obvious from analytic solutions alone.