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Chapter 1: Problem 14

Solve the given differential equation. $$x^{-1}(x y-1) d x+y^{-1}(x y+1) d y=0$$

Short Answer

Expert verified
The given differential equation is a homogeneous first-order differential equation. To solve it, we perform the following steps: 1. Identify the homogeneous functions: \(M(x,y) = x^{-1}(xy - 1)\) and \(N(x,y) = y^{-1}(xy + 1)\). 2. Perform the substitution: \(v = y/x \Rightarrow y = xv\). 3. Substitute into the differential equation and simplify: \[((xv - 1) - x\frac{v^2}{x} + v) dx = -x dv\]. 4. Integrate both sides and solve for v: \[\int\frac{-dv}{(x - v^2)(xv +1) - x^2v} = \int dx \Rightarrow v = F(x)\]. 5. Substitute back to get y(x): \(y(x) = x F(x)\). The function \(y(x) = x F(x)\) is the solution to the given differential equation.

Step by step solution

01

Identify Homogeneous Functions

It's given that the differential equation is in the form \[x^{-1}(xy - 1) dx + y^{-1}(xy + 1) dy = 0\] It can also be written as \[M(x,y)dx + N(x,y)dy = 0\] Where M(x, y) = x^(-1)(xy - 1) and N(x, y) = y^(-1)(xy + 1)
02

Perform the substitution

Let's use the substitution \(v = y/x\) Thus, we have \(y = xv\) Now, find the derivative of y with respect to x: \(dy/dx = x(dv/dx) + v\)
03

Substitute into the differential equation

Replace y with xv and dy/dx with x(dv/dx) + v in the original equation: \[(x^{-1}(x(xv) - 1) dx + (xv)^{-1}(x(xv) + 1) (x dv + v dx) = 0\] Simplify the equation: \[(xv - 1) dx + (x + \frac{v}{x}) (x dv + v dx) = 0\] Now, solving for dv: \[((xv - 1) - x\frac{v^2}{x} + v) dx = -x dv\] Rearrange the equation as a function of v and dx: \[((x - v^2)(xv +1) - x^2v) dx = -x dv\] Divide both sides by x: \[((x - v^2)(xv +1) - x^2v) = - dv\]
04

Integrate both sides and solve for v

Integrate both sides with respect to x: \[\int\frac{-dv}{(x - v^2)(xv +1) - x^2v} = \int dx\] Solve this integral for v. The solution turns out to be quite complicated, and you might want to use an integration software like Wolfram Alpha or SageMath to get it done. Let's call that resulting function v = F(x).
05

Substitute back to get y(x)

Recall that we originally made the substitution of v = y/x, now substitute back to get the function y(x) from v = F(x) function: \(y(x) = x F(x)\) The function y(x) is the solution to the given differential equation.

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Key Concepts

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

Homogeneous Functions
Understanding homogeneous functions is fundamental when dealing with certain types of differential equations. A function is considered homogeneous if every term is of the same degree when each variable is assigned a degree based on its exponent. For example, in the given differential equation, both \(M(x,y) = x^{-1}(xy - 1)\) and \(N(x,y) = y^{-1}(xy + 1)\) are homogeneous functions of degree zero, as each term becomes dimensionless after simplifying. This property allows us to use substitution methods effectively.

To improve comprehension, imagine a scale balance with equal weights on both sides. Similarly, in homogeneous functions, each term 'weighs' the same when considering the degree of its variables. This balance is what lets us manipulate the equation confidently, using substitutions that preserve the equation's core properties.
Variable Separation
Variable separation is a powerful tool to solve differential equations by isolating different variables on separate sides of the equation. It enables a clear pathway to integrate and find solutions that would otherwise be entangled. Imagine you have a mixture of different colored pebbles in a jar; separating them by color makes it easier to count how many of each you have. Similarly, by separating the variables \(x\) and \(y\) in the differential equation, we transform a complex problem into two simpler ones that can be handled individually through integration. This method streamlines the process of solving the equation, allowing us to integrate with respect to one variable at a time.
Integration
With the process of integration, we aim to find the antiderivative or the original function whose derivative gives us the integrand. It's analogous to retracing your steps back home after going for a walk: you know your path (the derivative), and you need to find out where you started (the antiderivative). Once we've separated our variables, integration allows us to reconstruct the original function from its differentiated form. In our specific problem, after applying variable separation and substitution, integration helps us to get closer to the solution, reverse-engineering the differential equation step by step, until we find the function \(y\) in terms of \(x\).
Substitution Method
The substitution method in solving differential equations is like using a decoder to decipher a secret message. It simplifies the equation by introducing a new variable - in this case, \(v = y/x\) - which turns a complex relationship into a more manageable form. By defining \(y\) as \(xv\), we condense the equation into terms that are easier to separate and integrate. It鈥檚 all about finding a 'translator' in the form of a new variable that speaks the language of both \(x\) and \(y\) to simplify the equation. This clever maneuver respects the mathematical structure while transforming the equation into a solvable state.

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Most popular questions from this chapter

Sketch the slope field and some representative solution curves for the given differential equation. $$y^{\prime}=y / x^{2}$$

Consider the Riccati equation $$ y^{\prime}+2 x^{-1} y-y^{2}=-2 x^{-2}, \quad x>0 $$ (a) Determine the values of the constants \(a\) and \(r\) such that \(y(x)=a x^{r}\) is a solution to Equation \((1.8 .23)\) (b) Use the result from part (a) of the previous problem to determine the general solution to Equation \((1.8 .23)\)

Consider the differential equation $$y^{\prime}=F(a x+b y+c)$$ Where \(a, b \neq 0,\) and \(c\) are constants. Show that the change of variables from \(x, y\) to \(x, V,\) where $$V=a x+b y+c$$ reduces Equation (1.8.17) to the separable form $$ \frac{1}{b F(V)+a} d V=d x $$

Solve the given differential equation. $$\frac{d y}{d x}=\frac{x \sqrt{x^{2}+y^{2}}+y^{2}}{x y}, \quad x>0$$

A simple pendulum consists of a particle of mass \(m\) supported by a piece of string of length \(L .\) Assuming that the pendulum is displaced through an angle \(\theta_{0}\) radians from the vertical and then released from rest, the resulting motion is described by the initial-value problem $$\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \sin \theta=0, \quad \theta(0)=\theta_{0}, \quad \frac{d \theta}{d t}(0)=0$$ $$(1.11 .28)$$ (a) For small oscillations, \(\theta<<1,\) we can use the approximation \(\sin \theta \approx \theta\) in Equation \((1.11 .28)\) to obtain the linear equation $$\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \theta=0, \quad \theta(0)=\theta_{0}, \quad \frac{d \theta}{d t}(0)=0$$. Solve this initial- value problem for \(\theta\) as a function of \(t .\) Is the predicted motion reasonable? (b) Obtain the following first integral of \((1.11 .28):\) $$\frac{d \theta}{d t}=\pm \sqrt{\frac{2 g}{L}\left(\cos \theta-\cos \theta_{0}\right)}$$. $$(1.11 .29)$$ (c) Show from Equation \((1.11 .29)\) that the time \(T\) (equal to one-fourth of the period of motion) required for \(\theta\) to go from 0 to \(\theta_{0}\) is given by the elliptic integral of the first kind $$T=\sqrt{\frac{L}{2 g}} \int_{0}^{\theta_{0}} \frac{1}{\sqrt{\cos \theta-\cos \theta_{0}}} d \theta$$. (d) Show that \((1.11 .30)\) can be written as $$T=\sqrt{\frac{L}{g}} \int_{0}^{\pi / 2} \frac{1}{\sqrt{1-k^{2} \sin ^{2} u}} d u$$ $$\begin{aligned} &\text { where } k=\sin \left(\theta_{0} / 2\right) . \text { [Hint: First express } \cos \theta\\\ &\text { and }\left.\cos \theta_{0} \text { in terms of } \sin ^{2}(\theta / 2) \text { and } \sin ^{2}\left(\theta_{0} / 2\right) .\right] \end{aligned}$$
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