/*! 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 11 Find the general solution of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general solution of the indicated differential equation. If possible, find an explicit solution. $$ y^{3} y^{\prime}=x+2 y^{\prime} $$

Short Answer

Expert verified
General solution: \(y^4 - 8y = 2x^2 + C\), no explicit solution unless further assumptions simplify.

Step by step solution

01

Separate Variables

Start by rewriting the given differential equation to separate variables. Given: \(y^{3} y^{\prime} = x + 2 y^{\prime}\). Rearrange to isolate \(y^{\prime}\) terms: \(y^3 y^{\prime} - 2 y^{\prime} = x\), which simplifies to \(y^{\prime}(y^3 - 2) = x\). Thus, \(y^{\prime} = \frac{x}{y^3 - 2}\).
02

Integrate Both Sides

Now integrate both sides with respect to \(x\). The left side becomes \( \int dy = \int \frac{x}{y^3 - 2} dx \). Separate variables again: \(y^3 - 2\) on the opposite side from \(dy\), \(\int (y^3 - 2) dy = \int x dx\). This gives \( \int x dx = \frac{x^2}{2}\) and the left requires rewriting if needed for complexity. For simplicity here assume direct solutions.
03

Solve the Integral

The integral \( \int (y^3 - 2) dy \) often needs substitution if complex, solved simplistically here assuming basic direct integration leads to: \(\int y^3 dy - \int 2 dy = \frac{y^4}{4} - 2y\). Recognize this integration aspect is critical for explicit steps.
04

General Solution

Combine integrated results: \(\frac{y^4}{4} - 2y = \frac{x^2}{2} + C\). This represents an implicit solution for \(y\) in terms of \(x\).
05

Check for Explicit Solution

Attempt to isolate \(y\) for explicit solution. Rearange: \(y^4 - 8y = 2x^2 + 8C\). Consider cases solving for \(y\) directly depending complexity. For ease, algebraically simplifying may not yield explicit easy solutions unless specific further simplification or assumptions considered.

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
Separation of Variables is a valuable method used in solving differential equations. It helps to simplify an equation so that each variable is isolated on different sides of the equation.
In our example, we started with the differential equation
  • \( y^{3} y^{\prime} = x + 2 y^{\prime} \)
The goal was to bring all terms involving \( y \) to one side and terms involving \( x \) to the other side.
By rearranging the equation, we got \( y^{\prime}(y^3 - 2) = x \) which allowed us to express \( y^{\prime} \) solely in terms of \( x \), such as \( y^{\prime} = \frac{x}{y^3 - 2} \).
This step is critical because it transforms the differential equation into a form that is more manageable for integration.
Integration
Integration is the mathematical process of finding the antiderivative of a function, and it plays a key role in solving differential equations. Once we separated variables, the next step was to integrate both sides of the equation.
We needed to integrate the expression \( \int (y^3 - 2) dy = \int x dx \). This is how the separated form looked after setting it up for integration.
The right side is straightforward, resulting in the expression \( \frac{x^2}{2} \) after integration. The left side was solved by recognizing it as \( \int y^3 dy - \int 2 dy \), which simplifies to \( \frac{y^4}{4} - 2y \).
Integration involves understanding and applying antiderivative formulas, which can often be straightforward but occasionally require substitution or more complex methods.
General Solution
The general solution of a differential equation is an equation that involves arbitrary constants, representing a family of curves.
After completing the integration, we arrived at the implicit solution:
  • \( \frac{y^4}{4} - 2y = \frac{x^2}{2} + C \)
Here, \( C \) is the arbitrary constant resulting from the integration process.
This equation embodies the general solution, where each different value of \( C \) corresponds to a different curve of solutions based on initial or boundary conditions provided in different scenarios.
Understanding the general solution highlights how one differential equation can have multiple solutions until additional information is provided.
Explicit Solution
An explicit solution presents the dependent variable, in this case \( y \), clearly isolated on one side of the equation.
To find an explicit solution, we try to express \( y \) entirely in terms of \( x \). Beginning with the equation
  • \( y^4 - 8y = 2x^2 + 8C \)
Isolating \( y \) explicitly is often desirable but can sometimes be complex, especially if the equation does not easily simplify.
In practice, explicit solutions might not always be possible without making additional assumptions or using numerical approaches.
Explicit solutions are advantageous as they provide a clearer functional form, aiding in further analysis or applications of the differential equation.

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

Is it possible to find a function \(f(t, x)\) that is continuous and has continuous partial derivatives such that the functions \(x_{1}(t)=t\) and \(x_{2}(t)=\sin t\) are both solutions to \(x^{\prime}=f(t, x)\) near \(t=0\) ?

Uniqueness is not just an abstraction designed to please theoretical mathematicians. For example, consider a cylindrical drum filled with water. A circular drain is opened at the bottom of the drum and the water is allowed to pour out. Imagine that you come upon the scene and witness an empty drum. You have no idea how long the drum has been empty. Is it possible for you to determine when the drum was full? (a) Using physical intuition only, sketch several possible graphs of the height of the water in the drum versus time. Be sure to mark the time that you appeared on the scene on your graph. (b) It is reasonable to expect that the speed at which the water leaves through the drain depends upon the height of the water in the drum. Indeed, Torricelli's law predicts that this speed is related to the height by the formula \(v^{2}=2 g h\), where \(g\) is the acceleration due to gravity near the surface of the earth. Let \(A\) and \(a\) represent the area of a cross section of the drum and drain, respectively. Argue that \(A \Delta=a v \Delta t\), and in the limit, \(A d h / d t=a v\). Show that \(d h / d t=-(a / A) \sqrt{2 g h}\). (c) By introducing the dimensionless variables \(\omega=\alpha h\) and \(s=\beta t\) and then choosing parameters $$ \alpha=\frac{1}{h_{0}} \quad \text { and } \quad \beta=\left(\frac{a}{A}\right) \sqrt{\frac{2 g}{h_{0}}}, $$ where \(h_{0}\) represents the height of a full tank, show that the equation \(d h / d t=-(a / A) \sqrt{2 g h}\) becomes \(d w / d s=-\sqrt{w}\). Note that when \(w=0\), the tank is empty, and when \(w=1\), the tank is full. (d) You come along at time \(s=s_{0}\) and note that the tank is empty. Show that the initial value problem, \(d w / d s=-\sqrt{w}\), where \(w\left(s_{0}\right)=0\), has an infinite number of solutions. Why doesn't this fact contradict the uniqueness theorem? Hint: The equation is separable and the graphs you drew in part (a) should provide the necessary hint on how to proceed.

Suppose that \(x\) is a solution to the initial value problem $$ x^{\prime}=x-t^{2}+2 t \text { and } x(0)=1 . $$ Show that \(x(t)>t^{2}\) for all \(t\) for which \(x\) is defined.

Calculate the differential \(d F\) for the given function \(F\). $$ F(x, y)=2 x y+y^{2} $$

Suppose that \(y\) is a solution to the initial value problem $$ y^{\prime}=y^{2}-\cos ^{2} t-\sin t \quad \text { and } \quad y(0)=2 $$ Show that \(y(t)>\cos t\) for all \(t\) for which \(y\) is defined.

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.