/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 In Problems 9-16, solve the give... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 15

In Problems 9-16, solve the given differential equation. $$ \left(x^{2}+4\right) d y=(2 x-8 x y) d x $$

Short Answer

Expert verified
Separate variables, integrate, and solve for \(y\). Solution: \(y = \frac{1}{4} (1 - K (x^2 + 4)^{-4})\).

Step by step solution

01

Separate Variables

The given differential equation is \((x^2 + 4)\, dy = (2x - 8xy)\, dx\). Our goal is to separate the variables such that terms involving \(y\) are on one side and terms involving \(x\) are on the other side. Divide both sides by \((x^2 + 4)\) and \((1 - 4y)\):\[\frac{dy}{1 - 4y} = \frac{2x}{x^2 + 4} dx.\]
02

Integrate Both Sides

Now, integrate both sides to find the general solution. Integrate the left side with respect to \(y\) and the right side with respect to \(x\):\[\int \frac{1}{1 - 4y} dy = \int \frac{2x}{x^2 + 4} dx.\]
03

Integrate the Left Side

For the left side, the integral is:\[-\frac{1}{4} \ln|1 - 4y| + C_1.\]
04

Integrate the Right Side

For the right side, use a substitution method. Let \(u = x^2 + 4\), then \(du = 2x dx\). Thus, the integral becomes:\[ \int \frac{du}{u} = \ln|u| + C_2.\]Substitute back the original variable:\[ \ln|x^2 + 4| + C_2.\]
05

Combine Integrals

The equation from integrating both sides becomes:\[-\frac{1}{4} \ln|1 - 4y| = \ln|x^2 + 4| + C.\]Where \(C = C_2 - C_1\).
06

Solve for the General Solution if Possible

To find \(y\) in terms of \(x\), solve the equation\[-\frac{1}{4} \ln|1 - 4y| = \ln|x^2 + 4| + C.\]Exponentiate both sides to eliminate the logarithm:\[ |1 - 4y|^{1/4} = rac{1}{(x^2 + 4) e^C}.\]
07

Simplify the General Solution

Solving for \(y\), we get:\[1 - 4y = (x^2 + 4)^{-4} e^{-4C} = K (x^2 + 4)^{-4},\] where \(K = e^{-4C}\).Thus, the solution is\[y = \frac{1}{4} (1 - K (x^2 + 4)^{-4}).\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Variable Separation
Differential equations can often be solved by the technique of "variable separation". This involves rearranging the equation so that all terms containing one variable appear on one side of the equation and all terms containing the other variable appear on the opposite side. In the given equation
  • The terms involving the variable \(y\) and \(dy\) must be separated from those involving \(x\) and \(dx\).
  • To do this, divide both sides of the equation by any factors necessary to isolate \(dy\) and \(dx\).
  • This results in a format where integration can occur with respect to each variable independently.
In our example: \[\frac{dy}{1 - 4y} = \frac{2x}{x^2 + 4} dx.\]This equation effectively separates the two variables, allowing the next step of integration to be tackled more easily.
Integration Techniques
When you have a differential equation that has been separated into distinct variable groups, the next step is integrating both sides. Integration techniques can vary based on the complexity of each side:
  • Left Side: For integrals like \(\int \frac{1}{1 - 4y} \, dy\), we recognize a standard form that directly relates to the natural logarithm, giving us: \[-\frac{1}{4} \ln |1 - 4y| + C_1.\]
  • Right Side: For \(\int \frac{2x}{x^2 + 4} \, dx\), we use substitution. Here, set \(u = x^2 + 4\), resulting in \( du = 2x \, dx\). This transforms the integral into: \[\int \frac{du}{u} = \ln |u| + C_2.\]
These methods effectively simplify complex expressions into more manageable functions that can be integrated with ease. Recognizing common forms or using substitutions is crucial to solving integrals.
General Solution Finding
After integrating both sides, the task is to solve for \(y\) to get the general solution of the differential equation. Begin with combining the results from both integrals:\[-\frac{1}{4} \ln |1 - 4y| = \ln |x^2 + 4| + C.\]The constant \(C\) represents an integration constant that can encapsulate any difference between \(C_1\) and \(C_2\). Then, to remove the logarithmic terms, exponentiate both sides:
  • This yields: \(|1 - 4y|^{1/4} = \frac{1}{(x^2 + 4) e^C}.\)
  • Solving for \(y\), you find:\[y = \frac{1}{4}(1 - K(x^2 + 4)^{-4}),\]
  • where \(K = e^{-4C}\) represents a constant derived from the integration process.
This final expression is your general solution, expressing \(y\) in a function of \(x\) and encompassing all particular solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) In Examples 3 and 4 of Section 2.1, we saw that any solution \(P(t)\) of \((4)\) possesses the asymptotic behavior \(P(t) \rightarrow a / b\) as \(t \rightarrow \infty\) for \(P_{0}>a / b\) and for \(0

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \frac{d r}{d \theta}+r \sec \theta=\cos \theta $$

Solve the given initial-value problem. $$ x^{2} \frac{d y}{d x}-2 x y=3 y^{4}, \quad y(1)=\frac{1}{2} $$

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \frac{d r}{d \theta}+r \sec \theta=\cos \theta $$

Heart Pacemaker A heart pacemaker consists of a switch, a battery of constant voltage \(E_{0}\), a capacitor with constant capacitance \(C\), and the heart as a resistor with constant resistance \(R\). When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage \(E\) across the heart satisfies the linear differential equation $$ \frac{d E}{d t}=-\frac{1}{R C} E $$ Solve the \(\mathrm{DE}\) subject to \(E(4)=E_{0}\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Space Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Oscillations

Read Explanation

Engineering Physics

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.