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Chapter 9: Problem 7

\(5-14\) Solve the differential equation. $$x y^{\prime}-2 y=x^{2}$$

Short Answer

Expert verified
The solution is \( y = x^2(\ln|x| + C) \).

Step by step solution

01

Identify the Type of Differential Equation

Given differential equation is \( x y' - 2y = x^2 \). This is a first-order linear differential equation of the form \( a(x) y' + b(x) y = c(x) \) where \( a(x) = x \), \( b(x) = -2 \), and \( c(x) = x^2 \).
02

Rewrite in Standard Form

Rewrite the equation in the standard form:\[ y' - \frac{2}{x} y = x \]
03

Find the Integrating Factor

The integrating factor \( \, \mu(x) \) is given by:\[ \mu(x) = e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln|x|} = |x|^{-2} \]Since we aren't considering \(x=0\) for differential equations, \(\mu(x) = x^{-2}\).
04

Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor:\[ x^{-2} y' - 2x^{-3} y = x^{-1} \]
05

Write the Left Side as a Derivative

Recognize that the left-hand side of the equation can now be written as the derivative:\[ \frac{d}{dx}(x^{-2} y) = x^{-1} \]
06

Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(x^{-2} y) \, dx = \int x^{-1} \, dx \] This gives:\[ x^{-2} y = \ln|x| + C \]
07

Solve for y

Multiply both sides by \(x^2\) to solve for \(y\):\[ y = x^2(\ln|x| + C) \] This is the general solution to the differential equation.

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Key Concepts

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

First-Order Linear Differential Equation
Let鈥檚 dive into what a first-order linear differential equation is. Essentially, it involves an unknown function and its derivative where the derivative is only raised to the first power. It typically takes the form: \[ a(x) y' + b(x) y = c(x) \] where:
  • \( y \) is the unknown function of \( x \)
  • \( y' \) is its derivative (rate of change)
  • \( a(x), b(x), \) and \( c(x) \) are functions of \( x \)
These equations can usually be solved using various techniques, one of which involves an integrating factor. The goal here is to find the unknown function \( y \) that satisfies the equation. As complex as it may initially seem, breaking it down into smaller parts helps make the process manageable. So, if you transform your given equation into a standard form, solving it becomes a systematic exercise.
Integrating Factor
Finding an integrating factor is a clever trick in differential equations. It allows us to simplify a differential equation into a form that's easier to solve. In our example, we have transformed the equation into a standard form a linear differential equation:\[ y' - \frac{2}{x} y = x \] The integrating factor, commonly denoted by \( \mu(x) \), is derived by:\[ \mu(x) = e^{\int b(x) \, dx} \] This factor turns the left-hand side of the equation into an easily integrable product of derivatives. In our original problem, the integrating factor simplifies to:\[ \mu(x) = e^{-2 \ln|x|} = |x|^{-2} \] Always remember that this step is crucial to transforming the equation into one that's solvable by integration. Once applied, it makes manipulating and integrating both sides to find \( y \) a straightforward process.
General Solution
The general solution of a differential equation provides a family of functions that satisfy the equation. This solution reflects not just one specific function, but a whole set. It accounts for any arbitrary constant "C", which stands as a placeholder for multiple possible solutions. In our exercise, after applying the integrating factor and integrating both sides, the equation simplifies to:\[ x^{-2} y = \ln|x| + C \]To isolate \( y \), multiply through by \( x^2 \):\[ y = x^2(\ln|x| + C) \] Here, \( C \) represents any constant value, meaning for each \( C \), there's a different specific solution. This whole range of solutions generated through varying \( C \) is essential for addressing initial value problems where starting conditions dictate the exact solution out of this family. Understanding the general solution allows you to tailor it to fit specific constraints based on real-world scenarios or initial conditions provided.

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Most popular questions from this chapter

For each predator-prey system, determine which of the variables, \(x\) or \(y,\) represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=-0.05 x+0.0001 x y \\\ \frac{d y}{d t} &=0.1 y-0.005 x y \\ \text { (b) } \frac{d x}{d t} &=0.2 x-0.0002 x^{2}-0.006 x y \\ \frac{d y}{d t} &=-0.015 y+0.00008 x y \end{aligned}$$

A vat with 500 gallons of beer contains 4\(\%\) alcohol (by volume). Beer with 6\(\%\) alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?

Consider the differential equation $$\frac{d P}{d t}=0.08 P\left(1-\frac{P}{1000}\right)-c$$ as a model for a fish population, where \(t\) is measured in weeks and \(c\) is a constant. (a) Use a CAS to draw direction fields for various values of \(c .\) (b) From your direction fields in part (a), determine the values of \(c\) for which there is at least one equilibrium solution. For what values of \(c\) does the fish population always die out? (c) Use the differential equation to prove what you dis- covered graphically in part (b). (d) What would you recommend for a limit to the weekly catch of this fish population?

\(9-10\) Sketch a direction field for the differential equation. Then use it to sketch three solution curves. $$y^{\prime}=x^{2}-y^{2}$$

Each system of differential equations is a model for two species that either compete for the same resources or cooperate for mutual benefit (flowering plants and insect pollinators, for instance). Decide whether each system describes competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=0.12 x-0.0006 x^{2}+0.00001 x y \\ \frac{d y}{d t} &=0.08 x+0.00004 x y \end{aligned}$$ $$\begin{aligned} \text { (b) } \frac{d x}{d t} &=0.15 x-0.0002 x^{2}-0.0006 x y \\ \frac{d y}{d t} &=0.2 y-0.00008 y^{2}-0.0002 x y \end{aligned}$$
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