/*! 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 13 Find general solutions of the di... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ x y^{\prime}=y+\sqrt{x^{2}+y^{2}} $$

Short Answer

Expert verified
The general solution is given implicitly by \( y = x(Cx - \sqrt{x^2 + y^2}) \).

Step by step solution

01

Rewrite the Differential Equation

Starting with the equation \( x y' = y + \sqrt{x^2 + y^2} \), we can first aim to rewrite it for clarity. Express it as \( x \frac{dy}{dx} = y + \sqrt{x^2 + y^2} \) and solve for \( \frac{dy}{dx} \). Divide the entire equation by \( x \) to isolate the derivative: \( \frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} \).
02

Introduce Substitution

To simplify, introduce a substitution. Let \( y = vx \), hence \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) into the differential equation: \((x) (v + x\frac{dv}{dx}) = vx + \sqrt{x^2 + (vx)^2}\).
03

Simplify with the Substitution

The equation becomes \(xv + x^2 \frac{dv}{dx} = vx + \sqrt{x^2 + v^2x^2} = vx + x\sqrt{1 + v^2}\). Simplifying: \(xv + x^2 \frac{dv}{dx} = vx + x\sqrt{1+v^2}\). Cancel \(vx\) from both sides: \(x^2 \frac{dv}{dx} = x\sqrt{1 + v^2}\). Divide by \( x \): \(x \frac{dv}{dx} = \sqrt{1 + v^2}\).
04

Separate Variables

Separate variables by writing \( \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \). Integrating both sides gives \( \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x} \).
05

Integrate Both Sides

The left side, \( \int \frac{dv}{\sqrt{1+v^2}} \), integrates to \( \ln|v + \sqrt{1+v^2}| \). The right side, \( \int \frac{dx}{x} \), integrates to \( \ln|x| \). Thus, \( \ln|v + \sqrt{1+v^2}| = \ln|x| + C \).
06

Solve for v

Exponentiating both sides removes the logarithm: \( |v + \sqrt{1+v^2}| = C|x| \). Isolating \( v \), we consider \( v + \sqrt{1+v^2} = Cx \) (since both sides should be positive when exponentials are equated).
07

Back-Substitute for y

Recall \( v = \frac{y}{x} \). Substitute back to get \( \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = \frac{y}{x} + \sqrt{\frac{x^2+y^2}{x^2}} = Cx \). Simplifying results in \( y = x(Cx - \sqrt{x^2 + y^2}) \). Rewriting leads to the implicit solution in terms of variables.

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 91Ó°ÊÓ!

Key Concepts

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

Separation of Variables
The separation of variables is a technique used to solve differential equations where you can isolate each variable on different sides of the equation. This method is particularly useful for equations that can be expressed in the form \( f(y) \, dy = g(x) \, dx \). Once separated, both sides of the equation are integrated independently. Here’s how it works step-by-step:Starting from the equation given in the exercise, after substitution and simplification, you arrive at \( x \frac{dv}{dx} = \sqrt{1 + v^2} \). The next step is to separate terms involving \( v \) from those involving \( x \). You achieve this by rewriting it as \( \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} \).
  • Integrate both sides separately.
  • The left side integrates to \( \ln |v + \sqrt{1+v^2}| \), and the right side to \( \ln |x| \).
After integrating, you solve for \( v \) by exponentiating the result to eliminate the logarithms. This final step brings you closer to the solution by expressing one variable in terms of the other plus an integration constant.
Substitution Method
The substitution method is a powerful strategy to tackle complex differential equations by simplifying them into a more manageable form. The core idea is to transform the equation by replacing variables with new ones, often simplifying the differential equation’s structure.In the original exercise, the substitution \( y = vx \) is introduced. This clever choice reflects the fact that the original equation involves a relationship between \( y \) and \( x \). By expressing \( y \) as a product of \( v \) and \( x \), the derivative \( \frac{dy}{dx} \) simplifies to \( v + x \frac{dv}{dx} \).
  • This substitution helps simplify the terms into a form where separation of variables is applicable.
  • It reduces the complexity, making integration straight-forward.
Choosing the optimal substitution requires identifying if the original equation has terms which can be organized into a simpler form or solved directly after the substitution.
Homogeneous Equations
Homogeneous equations are a special class of differential equations. They are characterized by a specific structure in which all terms are proportional to a function of the same degree. Recognizing an equation as homogeneous allows you to apply specific techniques to simplify and solve it.In the exercise, the equation \( xy' = y + \sqrt{x^2 + y^2} \) is addressed by rewriting terms and making use of substitution. The form \( y = vx \) helps transform the original, more complex expression into a homogeneous representation.
  • With \( y = vx \), all terms in the equation become divisible by the degree of \( x \), simplifying the function.
  • This expression helps to visualize the equation in a form ready for techniques like separation of variables.
Homogeneous equations often allow elegance in solving, due to their consistent structure, revealing solutions that might not be evident from the original formulation.

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

Thousands of years ago ancestors of the Native American's crossed the Bering Strait from Asia and entered the western hemisphere. Since then, they have fanned out across North and South America. The single language that the original Native Americans spoke has since split into many Indian "language families." Assume (as in Problem 52 ) that the number of these language families has been multiplied by \(1.5\) every 6000 years. There are now 150 Native American language families in the western hemisphere. About when did the ancestors of today's Native Americans arrive?

Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ 2 x y^{\prime}+y^{3} e^{-2 x}=2 x y $$

Verify that the given differential equation is exact; then solve it. $$ \left(x^{3}+\frac{y}{x}\right) d x+\left(y^{2}+\ln x\right) d y=0 $$

The amount \(A(t)\) of atmospheric pollutants in a certain mountain valley grows naturally and is tripling every \(7.5\) years. (a) If the initial amount is 10 pu (pollutant units), write a formula for \(A(t)\) giving the amount (in pu) present after \(t\) years. (b) What will be the amount (in pu) of pollutants present in the valley atmosphere after 5 years? (c) If it will be dangerous to stay in the valley when the amount of pollutants reaches 100 pu, how long will this take?

Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ x y^{2} y^{\prime}=x^{3}+y^{3} $$

See all solutions

Recommended explanations on Math Textbooks

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.