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Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable. $$x u^{\prime}(x)=u^{2}-4$$

Short Answer

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Based on the step by step solution provided above, answer the following question: Q: Find the general solution of the given first-order, nonlinear, autonomous differential equation: $$x u^{\prime}(x)=u^{2}-4$$ A: The general solution of the given differential equation explicitly as a function of the independent variable x is: $$u(x) = \frac{2 \pm (2 + 2x^4e^{4C})}{1 - x^4e^{4C}}$$

Step by step solution

01

Rewrite the Differential Equation

First, let's rewrite the equation in a more convenient form to apply the separation of variables method. We have $$u^{\prime}(x)=\frac{u^{2}-4}{x}$$
02

Separate the Variables

Separate the variables u and x, to make the equation solvable, so we have: $$\frac{du}{u^{2}-4} = \frac{dx}{x}$$
03

Integrate Both Sides

Now, integrate both sides of the equation with respect to the separated variables: $$\int\frac{du}{u^{2}-4} = \int\frac{dx}{x}$$
04

Perform the Integration

Use partial fractions for the left side equation and logarithm for the right side: $$\frac{1}{4}\ln\left|\frac{u-2}{u+2}\right| = \ln|x| + C$$ Where C is the constant of integration.
05

Solve for u(x)

Now, we will solve the equation for u explicitly as a function of x: $$\ln\left|\frac{u-2}{u+2}\right|^{1/4} = \ln|x| + C$$ Raise both sides to the power of e (i.e. opposite of the natural logarithm) to get rid of the ln: $$\frac{u-2}{u+2} = \pm x^{4}e^{4C}$$ Now, we make u the subject of the formula: $$u(x) = \frac{2 \pm (2 + 2x^4e^{4C})}{1 - x^4e^{4C}}$$ Now we have obtained the general solution of the given equation explicitly as a function of the independent variable x: $$u(x) = \frac{2 \pm (2 + 2x^4e^{4C})}{1 - x^4e^{4C}}$$

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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 technique for solving differential equations, where we rearrange the equation to isolate one variable on each side.

Consider the example differential equation given by \( u'(x) = \frac{u^2 - 4}{x} \). To use separation of variables, we aim to express the differentials \( du \) and \( dx \) on opposite sides of the equality, as seen in the steps of the solution. The equation becomes \( \frac{du}{u^2 - 4} = \frac{dx}{x} \), which allows us to apply integration to both sides separately.

The beauty of this technique is that it simplifies many complex differential equations into a form where we can find solutions by integrating, as it decouples the variables and turns a potentially difficult differential equation into a simpler integral one.
Partial Fraction Decomposition
Partial fraction decomposition is an algebraic method used to simplify complex fractions into simpler parts that are easier to integrate.

In the problem \( \int\frac{du}{u^2 - 4} \), the denominator can be factored into \( (u - 2)(u + 2) \). This expression is not readily integrable, but by decomposing it into partial fractions, we get terms that can be integrated using simple integration techniques.

For instance, a fraction like \( \frac{1}{u^2 - 4} \) would be decomposed into \( \frac{A}{u - 2} + \frac{B}{u + 2} \) where A and B are constants determined through algebraic means. Once these constants are found, integration becomes straightforward. Partial fraction decomposition is especially useful when dealing with rational expressions, and it is often employed before integration to make the process more manageable.
Integration Techniques
Integration techniques are numerous and varied, designed to address the multitude of forms that functions can take. From basic antiderivatives to more complex methods such as integration by parts, each technique serves to find the integral of functions that cannot be integrated directly.

For instance, in our exercise, the right side of our separated equation, \( \int\frac{dx}{x} \), involves a simple logarithmic integration that results in \( \ln|x| \). On the other hand, the left side of the equation requires partial fraction decomposition before carrying out the integration because the denominator is a quadratic expression. Knowledge of integration techniques is crucial when handling different kinds of functions, as it equips you with the right method to find the integral effectively.
Natural Logarithm Properties
The natural logarithm properties can simplify complex expressions and solve equations involving logarithms. The function \( \ln(x) \) is the inverse of the exponential function \( e^x \), and they cancel each other out when applied successively.

In the solution, we have \( \frac{1}{4}\ln\left|\frac{u-2}{u+2}\right| = \ln|x| + C \), which represents the antiderivative of both sides of our equation. We exploit the logarithmic property that \( \ln(a^b) = b \ln(a) \) to raise the logarithm out of the exponent, and we later use the exponential function to 'cancel out' the logarithm. By doing this, we get rid of the logarithm and solve for \( u \) explicitly in terms of \( x \) and the constant \( C \), resulting in the general solution of the differential equation. Understanding natural logarithm properties is crucial in integrating functions that result in logarithmic forms.

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Most popular questions from this chapter

A fish hatchery has 500 fish at \(t=0,\) when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b\) \(y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0,\) in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A one-million-liter pond is contaminated by a chemical pollutant with a concentration of \(20 \mathrm{g} / \mathrm{L} .\) The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr}\). Assuming the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$t y^{\prime}(t)+y=1+t, y(1)=4$$

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\). a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that \(\frac{d y}{d x}=\frac{-2 x}{y}\) b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{c}|x|\) and then explain why it follows that \(y^{2}=k x\) where \(k\) is an arbitrary constant. Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\).

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{1}{t} y(t)=0, y(1)=6$$

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