Problem 32 - Advanced Engineering Mathematics

Question

The differential equation \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{2}-2 x\right) / y\) can be solved by separating variables. It follows that \(y^{2} / 2=\left(x^{3} / 3\right)-x^{2}+c\) and \(\operatorname{since} \mathbf{X}(0)=\left(x(0), x^{\prime}(0)\right)=(1,0), c=\frac{2}{3} .\) Therefore The differential equation \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{2}-2 x\right) / y\) can be solved by separating variables. It follows that \(y^{2} / 2=\left(x^{3} / 3\right)-x^{2}+c\) and \(\operatorname{since} \mathbf{X}(0)=\left(x(0), x^{\prime}(0)\right)=(1,0), c=\frac{2}{3} .\) Therefore $$\frac{y^{2}}{2}=\frac{x^{3}-3 x^{2}+2}{3}=\frac{(x-1)\left(x^{2}-2 x-2\right)}{3}$$ But \((x-1)\left(x^{2}-2 x-2\right)>0\) for \(1-\sqrt{3} < x < 1\) and so each \(x\) in this interval has 2 corresponding values of \(y\) therefore \(\mathbf{X}(t)\) is a periodic solution.

Step-by-Step Solution

Answer

The solution is periodic for \( 1 - \sqrt{3} < x < 1 \) with two y-values for each x.

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{x^2 - 2x}{y} \). Our goal is to find the function \( y(x) \) by separating the variables.

Step 2: Separate Variables

Rearrange the differential equation to separate the variables: \( y \, dy = (x^2 - 2x) \, dx \). This allows us to integrate both sides with regard to their respective variables.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[ \int y \, dy = \int (x^2 - 2x) \, dx \] Which results in: \[ \frac{y^2}{2} = \frac{x^3}{3} - x^2 + C \] where \( C \) is the constant of integration.

Step 4: Insert Initial Conditions and Solve for C

Given that initially \( x(0) = 1 \) and \( y(0) = 0 \), substitute these into the integrated equation to find the constant \( C \): \[ \frac{0^2}{2} = \frac{1^3}{3} - 1^2 + C \] Solving, we get \[ 0 = \frac{1}{3} - 1 + C \] \[ C = \frac{2}{3} \]

Step 5: Rewrite the Solution Equation with C Value

Substituting \( C = \frac{2}{3} \) back into the integrated equation: \[ \frac{y^2}{2} = \frac{x^3}{3} - x^2 + \frac{2}{3} \] This simplifies to: \[ \frac{y^2}{2} = \frac{x^3 - 3x^2 + 2}{3} \]

Step 6: Analyze the Solution

The equation \( \frac{(x-1)(x^2 - 2x - 2)}{3} \) shows that for \( 1 - \sqrt{3} < x < 1 \) each \( x \) has two corresponding \( y \)-values. Therefore, the solution \( \mathbf{X}(t) \) is periodic within this interval.

Key Concepts

Separation of Variables

Separation of variables is a common method used to solve differential equations. The basic idea is to rearrange a differential equation so that each variable is on a different side of the equation. In our original problem, the differential equation is given as \( \frac{dy}{dx} = \frac{x^2 - 2x}{y} \). By using the separation of variables, we rearrange the equation such that \( y \, dy = (x^2 - 2x) \, dx \).

This rearrangement allows us to integrate each side independently. Essentially, we are treating the derivatives as if they are fractions that can be separated and recombined after integration. This simplification is the key advantage of this method. After separating the variables, the next step is to integrate both sides, but each with respect to their corresponding variables. This process results in a more manageable form that we can solve to find the function \( y(x) \).

Constant of Integration

The constant of integration plays a crucial role in solving differential equations. When we integrate a function, the result will include a constant because integration is essentially the reverse of differentiation, and different functions can have the same derivative.

For the exercise, after integrating \( y \, dy = (x^2 - 2x) \, dx \), we get \( \frac{y^2}{2} = \frac{x^3}{3} - x^2 + C \). Here, \( C \) is the constant of integration.

The value of \( C \) is determined using initial conditions. In this case, we used the initial condition \( x(0) = 1 \) and \( y(0) = 0 \) which lead us to finding \( C = \frac{2}{3} \). Initial conditions are essential because they allow us to determine this constant, which then gives us a particular solution to the differential equation, rather than a family of solutions.

Periodic Solutions

Periodic solutions are solutions to differential equations that repeat at regular intervals. In our context, the solution to the differential equation \( \frac{dy}{dx} = \frac{x^2 - 2x}{y} \) becomes periodic under certain conditions. This periodic nature occurs when, within a certain interval, each \( x \) has two corresponding \( y \)-values that repeat after a fixed period.

In this problem, after solving and simplifying the integrated equation, we analyze that the solution for \( x \) in the interval \( 1 - \sqrt{3} < x < 1 \), shows this behavior. During this phase, the function literally repeats itself, making it a periodic solution.

Periodic solutions are often used in modeling real-world phenomena that are cycles, such as waves, vibrations, or other repetitive patterns, making them an important aspect to understand in the study of differential equations.