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Chapter 8: Problem 37

Find the general solution of the following differential equations. $$u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}$$

Short Answer

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Question: Find the general solution of the first-order differential equation: $$u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}$$ Answer: The general solution of the given differential equation is: $$u(x)=\frac{1}{2}\ln{\frac{|x^2+4|}{x^2+4}}+C$$

Step by step solution

01

Identify the type of differential equation

We have a first-order differential equation of the form: $$u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}$$ This is a first-order, separable differential equation.
02

Separate the variables

To separate the variables, we need to rewrite the equation in the form of \(u^{\prime}(x) = f(x)\). In this case, we have: $$u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}$$
03

Integrate both sides

Now we'll integrate both sides of the equation with respect to x: $$\int u^{\prime}(x) \, dx = \int \frac{2(x-1)}{x^{2}+4} \, dx$$ Let's focus on the integral on the right side.
04

Apply substitution to the integral

We'll perform a substitution to simplify the integral. Let \(t=x^2 + 4\). Then, \(dt = 2x \, dx\). So, the integral becomes: $$\int \frac{2(x-1)}{t}\frac{dt}{2x}= \int \frac{(x-1)}{t} \, dt$$
05

Complete the integration

Now, we can perform the integration: $$u(x) = \int \frac{(x-1)}{x^2+4}\, dx = \int \frac{(x-1)}{t} \, dt$$ Expand the fraction in the integral: $$u(x) = \int \frac{x}{t} \, dt - \int \frac{1}{t} \, dt$$ Here, perform the integration to get: $$u(x)=\frac{1}{2}\ln|x^2+4|-\frac{1}{2}\ln{t}+C$$
06

Substitute back and simplify

Swap back the substitution we made earlier, \(t = x^2 + 4\): $$u(x)=\frac{1}{2}\ln|x^2+4|-\frac{1}{2}\ln{x^2+4}+C$$ Combine the constants: $$u(x)=\frac{1}{2}\ln{\frac{|x^2+4|}{x^2+4}}+C$$ The general solution of the given differential equation is: $$u(x)=\frac{1}{2}\ln{\frac{|x^2+4|}{x^2+4}}+C$$

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Key Concepts

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

General Solution
A general solution to a differential equation is a formula that encompasses all possible solutions. It includes an arbitrary constant, often represented by "C", which allows for a continuity of solutions, each corresponding to different initial conditions. In this context, finding the general solution of the differential equation \[u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}\] involves crafting a function, denoted as \(u(x)\), that represents the behavior of the differential equation across a range of values.
The general solution provides a wide scope of possible outcomes based on varying 'C' values, enabling it to align with specific initial conditions you might encounter.
  • The constant \(C\) does not change the derivative, keeping the differential equation satisfied.
  • Solving for the general solution often requires integration, capturing every potential behavior of the differential we began with.
Variable Separation
Variable separation is a method used to solve first-order separable differential equations, where you can separate the variables on either side of the equation. This technique essentially allows you to rearrange the equation such that each variable and its differential coefficient is on separate sides.
In our example: \[u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}\]we begin by securing all terms involving \(x\) on one side, freeing \(u'(x)\) and relating it distinctly to its own variable.
This results in a setup that gives us an equation prime for integration.
  • Separation leads to \(u^{\prime}(x) \, dx = \frac{2(x-1)}{x^2+4}\, dx\), preparing for the next integration step.
  • Effectively separates an equation into distinct parts, making it simpler to solve.
Substitution Method
In tackling integrals, the substitution method is a handy tool that can simplify complex expressions. It involves replacing complicated parts of an integral with a substitution variable, which makes the integration process more manageable.
For the differential equation at hand, when integrating \[\int \frac{2(x-1)}{x^2+4} \, dx\], we can apply a substitution \(t = x^2 + 4\) which simplifies calculations.
This step changes the variables and simplifies the expression to something more straightforward, altering the integration path constructively.
  • Substitution helps in transforming a difficult integral into an easier one.
  • Makes it feasible to handle by introducing a new "t" variable to lighten computations.
Integration
Integration is the inverse operation of differentiation. It's an essential tool in finding the general solution of differential equations; in this context, it determines the behavior over an interval by summing infinitesimal changes.
By integrating the equation \[\int \frac{2(x-1)}{x^2+4} \, dx\], we accumulate areas and curves defined by our initial differential equation. Completing the integration here requires executing the substitution, yielding a more straightforward expression \[u(x)=\int \frac{(x-1)}{t} \, dt\].
Upon simplifying, it leads us finely to a final integrative expression: \[u(x) = \frac{1}{2}\ln{\left| x^2+4 \right|} + C\]
This solution includes an arbitrary constant \(C\), ensuring that it represents a wide spectrum of answers.
  • Integrating parameterizes the changes in \(x\) to assign values to \(u\).
  • Recognizes both symmetry and defined value changes across intervals.

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Most popular questions from this chapter

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. $$y^{\prime}(t)+\frac{2 t}{t^{2}+1} y(t)=1+3 t^{2}, \quad y(1)=4$$

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.

Analytical solution of the predator-prey equations The solution of the predator-prey equations $$x^{\prime}(t)=-a x+b x y, y^{\prime}(t)=c y-d x y$$ can be viewed as parametric equations that describe the solution curves. Assume that \(a, b, c,\) and \(d\) are positive constants and consider solutions in the first quadrant. a. Recalling that \(\frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)},\) divide the first equation by the second equation to obtain a separable differential equation in terms of \(x\) and \(y\) b. Show that the general solution can be written in the implicit form \(e^{d x+b y}=C x^{c} y^{a},\) where \(C\) is an arbitrary constant. c. Let \(a=0.8, b=0.4, c=0.9,\) and \(d=0.3 .\) Plot the solution curves for \(C=1.5,2,\) and \(2.5,\) and confirm that they are, in fact, closed curves. Use the graphing window \([0,9] \times[0,9]\)

Consider the following pairs of differential equations that model a predator- prey system with populations \(x\) and \(y .\) In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=-3 x+x y, y^{\prime}(t)=2 y-x y$$

Suppose an object with an initial temperature of \(T_{0}>0\) is put in surroundings with an ambient temperature of \(A\) where \(A<\frac{T_{0}}{2}\). Let \(t_{1 / 2}\) be the time required for the object to cool to \(\frac{T_{0}}{2}\) a. Show that \(t_{1 / 2}=-\frac{1}{k} \ln \left[\frac{T_{0}-2 A}{2\left(T_{0}-A\right)}\right]\) b. Does \(t_{1 / 2}\) increase or decrease as \(k\) increases? Explain. c. Why is the condition \(A<\frac{T_{0}}{2}\) needed?
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