Problem 38 - Advanced Engineering Mathematics

Question

If we let \(d x / d t=y,\) then \(d y / d t=-x^{3}-x .\) From this we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-\frac{x^{3}+x}{y}$$ Separating variables and integrating we obtain $$\int y d y=-\int\left(x^{3}+x\right) d x$$ and $$\frac{1}{2} y^{2}=-\frac{1}{4} x^{4}-\frac{1}{2} x^{2}+c_{1}$$ Completing the square we can write the solution as \(y^{2}=-\frac{1}{2}\left(x^{2}+1\right)^{2}+c_{2} .\) If \(\mathbf{X}(0)=\left(x_{0}, 0\right),\) then \(c_{2}=\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}\) and so $$\begin{aligned} y^{2} &=-\frac{1}{2}\left(x^{2}+1\right)^{2}+\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}=\frac{x_{0}^{4}+2 x_{0}^{2}+1-x^{4}-2 x^{2}-1}{2} \\ &=\frac{\left(x_{0}^{2}+x^{2}\right)\left(x_{0}^{2}-x^{2}\right)+2\left(x_{0}^{2}-x^{2}\right)}{2}=\frac{\left(x_{0}^{2}+x^{2}+2\right)\left(x_{0}^{2}-x^{2}\right)}{2} \end{aligned}$$ Note that \(y=0\) when \(x=-x_{0} .\) In addition, the right-hand side is positive for \(-x_{0}

Step-by-Step Solution

Answer

The solution is periodic with a center at (0,0) for the initial condition \(\mathbf{X}(0) = (x_0, 0)\).

Step 1: Understand the Given Differential Equations

You are given two differential equations: \( \frac{dx}{dt} = y \) and \( \frac{dy}{dt} = -x^3-x \). These describe how the variables \( x \) and \( y \) change with respect to time \( t \).

Step 2: Formulate the First-Order Differential Equation

You derive a first-order differential equation from the above by using \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{x^3 + x}{y} \), which expresses how \( y \) changes with respect to \( x \).

Step 3: Separate Variables and Integrate

Separate the variables by writing \( y \, dy = - (x^3 + x) \, dx \) and integrate both sides: \( \int y \, dy = - \int (x^3 + x) \, dx \).

Step 4: Evaluate the Integrals

The integration results in \( \frac{1}{2} y^2 = -\frac{1}{4} x^4 - \frac{1}{2} x^2 + c_1 \), where \( c_1 \) is a constant of integration.

Step 5: Complete the Square

Rewrite the expression in a completed square form: \( y^2 = -\frac{1}{2}(x^2 + 1)^2 + c_2 \), where \( c_2 \) is derived from \( c_1 \).

Step 6: Apply Initial Conditions

Using the initial condition \( \mathbf{X}(0) = (x_0, 0) \), deduce \( c_2 = \frac{1}{2}(x_0^2 + 1)^2 \).

Step 7: Express the Solution

Substitute \( c_2 \) back into the equation to get \( y^2 = \frac{(x_0^2 + x^2 + 2)(x_0^2 - x^2)}{2} \), representing a family of solutions based on initial \( x_0 \) values.

Step 8: Analyze the Nature of the Solution

Identify that \( y = 0 \) when \( x = -x_0 \) and deduce that the solution is periodic between \(-x_0\) and \(x_0\), with \( (0,0) \) as a center.

Key Concepts

Separation of Variables

Separation of variables is a powerful technique for solving differential equations. It transforms a complex problem into simpler integrals that are easier to evaluate. In the exercise above, the equation \(-\frac{x^3 + x}{y}\) emerges from rearranging the terms, leading us to express the derivative \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). This is set up by multiplying each side to isolate the variables: \(y \, dy = - (x^3 + x) \, dx\).

When employing separation of variables, the aim is to integrate both sides independently. Here we do this by integrating \(y \, dy\) and \(-(x^3 + x) \, dx\). These integrations are straightforward tools that provide a solution to the differential equation as shown:

• \(\int y \, dy = \frac{1}{2} y^2\)
• \(\int (x^3 + x) \, dx = \frac{1}{4} x^4 + \frac{1}{2} x^2\)

By separating variables and integrating, we discover the relationship between \(x\) and \(y\), giving rise to a clear form of the solution.

Periodic Solutions

A periodic solution refers to a solution that repeats values at regular intervals. In the context of differential equations, this typically means the function returns to the same state after a fixed time period or a range of input values.
In this exercise, the periodic nature emerges from the equation for \(y^2\) given by the completed square form: \[y^2 = \frac{(x_0^2 + x^2 + 2)(x_0^2 - x^2)}{2}\]Here we observe that the right-hand side becomes positive between \(-x_0\) and \(x_0\), suggesting that the values for \(y\) repeat across this interval, hence forming periodic behavior.

This characteristic identifies \((0,0)\) as a center, meaning it's a point that the solution cycles around. Periodic solutions are significant in many fields such as physics and engineering as they model cyclic phenomena like pendulums and oscillations.

Initial Conditions

Initial conditions are critical in solving differential equations as they determine specific solutions out of a broader family of potential solutions. The exercise provides initial conditions through \(\mathbf{X}(0) = (x_0, 0)\), guiding the determination of constants in the integrated expression.

Applying the initial condition means substituting the given initial value into your solution to determine the constant. In this case, \(c_2\) is calculated using \(c_2 = \frac{1}{2}(x_0^2 + 1)^2\), ensuring that the solution satisfies the initial condition.
Initial conditions function like GPS coordinates in the world of differential equations: they pinpoint the exact starting point, shaping the resulting path of the entire solution; thus, giving a clear, unique solution within a continuous set of functions. They ensure that the differential solutions match given real-world or theoretical scenarios accurately.