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Chapter 11: Problem 32

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.

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.
04

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 \[ 0 = \frac{1}{3} - 1 + C \] \[ C = \frac{2}{3} \]
05

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: \[ \frac{y^2}{2} = \frac{x^3 - 3x^2 + 2}{3} \]
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The differential equation is \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{3}-x\right) / y\) and so \(y^{2} / 2=x^{4} / 4-x^{2} / 2+c\) or \(y^{2}=x^{4} / 2-x^{2}+c_{1}\) since \(x(0)=0\) and \(y(0)=x^{\prime}(0)=v_{0},\) it follows that \(c_{1}=v_{0}^{2}\) and so $$y^{2}=\frac{1}{2} x^{4}-x^{2}+v_{0}^{2}=\frac{\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1}{2}$$The \(x\) -intercepts on this graph satisfy $$x^{2}=1 \pm \sqrt{1-2 v_{0}^{2}}$$ and so we must require that \(\left.1-2 v_{0}^{2} \geq 0 \text { (or }\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\right)\) for real solutions to exist. If \(x_{0}^{2}=1-\sqrt{1-2 v_{0}^{2}}\) and \(-x_{0} < x < x_{0},\) then \(\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1 > 0\) and so there are two corresponding values of \(y .\) Therefore \(\mathbf{X}(t)\) with \(\mathbf{X}(0)=\left(0, v_{0}\right)\) is periodic provided that \(\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\)

From \(\sin y=0\) we have \(y=\pm n \pi .\) From \(e^{x-y}=1,\) we can conclude that \(x-y=0\) or \(x=y .\) The critical points of the system are therefore \((\pm n \pi, \pm n \pi)\) for \(n=0,1,2, \ldots\)

We have \(d y / d x=y^{\prime} / x^{\prime}=-f(x) / y\) and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or } \quad y^{2}+2 F(x)=c$$ . We can conclude that for a given value of \(x\) there are at most two corresponding values of \(y .\) If (0,0) were a stable spiral point there would exist an \(x\) with more than two corresponding values of \(y .\) Note that the condition \(f(0)=0\) is required for (0,0) to be a critical point of the corresponding plane autonomous system \(x^{\prime}=y, y^{\prime}=-f(x)\).

(a) If \(f(x)=x^{2} / 2, f^{\prime}(x)=x\) and so $$\frac{d y}{d x}=\frac{y^{\prime}}{x^{\prime}}=-g \frac{x}{1+x^{2}} \frac{1}{y}$$. We can separate variables to show that \(y^{2}=-g \ln \left(1+x^{2}\right)+c .\) But \(x(0)=x_{0}\) and \(y(0)=x^{\prime}(0)=v_{0}\) Therefore \(c=v_{0}^{2}+g \ln \left(1+x_{0}^{2}\right)\) and so $$y^{2}=v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right)$$. Now $$v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right) \geq 0 \quad \text { if and only if } \quad x^{2} \leq e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1$$. Therefore, if \(|x| \leq\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2},\) there are two values of \(y\) for a given value of \(x\) and so the solution is periodic. (b) since \(z=x^{2} / 2,\) the maximum height occurs at the largest value of \(x\) on the cycle. From (a), \(x_{\max }=\) \(\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2}\) and so $$z_{\max }=\frac{x_{\max }^{2}}{2}=\frac{1}{2}\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]$$.

(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.
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