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Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x .\) \frac{d y}{d x}=\left(1+y^{2}\right) x^{2}

Short Answer

Expert verified
The family of solutions is \( y = \tan \left( \frac{x^3}{3} + C \right) \).

Step by step solution

01

Rewrite the differential equation

The given differential equation is \( \frac{dy}{dx} = (1+y^2)x^2 \). We want to separate the variables, so we'll rewrite it as \( dy = (1+y^2)x^2 dx \).
02

Separate the variables

To separate the variables \( y \) and \( x \), we divide both sides by \( 1+y^2 \): \( \frac{dy}{1+y^2} = x^2 dx \).
03

Integrate both sides

Integrate both sides separately. The left side integrates to \( \arctan(y) \), and the right side integrates to \( \frac{x^3}{3} + C \), where \( C \) is the constant of integration. Therefore, \( \arctan(y) = \frac{x^3}{3} + C \).
04

Solve for \( y \)

We'll solve for \( y \) to express \( y \) explicitly as a function of \( x \). Taking the tangent of both sides, we get \( y = \tan \left( \frac{x^3}{3} + C \right) \). This represents the family of solutions.

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Key Concepts

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

Separation of Variables
To tackle a differential equation like the one given, we often use the method of separation of variables. This technique simplifies solving differential equations when variables can be isolated onto different sides of the equation.
The primary goal of this method is to rewrite the differential equation such that all terms involving the dependent variable (in our case, \(y\)) are on one side and all terms involving the independent variable (\(x\)) are on the other side.
  • First, identify the parts of the equation that relate to \(y\) and \(x\).
  • Rewrite the equation so that the \(dy\) terms are on one side and the \(dx\) terms are on the other.
In the given problem, we started with the differential equation \( \frac{dy}{dx} = (1+y^2)x^2 \). By rewriting it as \( dy = (1+y^2)x^2 dx \) and then dividing by \(1+y^2\), the equation becomes \( \frac{dy}{1+y^2} = x^2 dx \). This separation of variables allows each side of the equation to be integrated individually.
Integration
Once the variables are separated, we proceed to integrate both sides. Integration is a mathematical process that helps us find the function from its derivative.
The idea is to calculate an antiderivative or the original function from the rate of change, which we have expressed with differentials.
  • The left side: \( \int \frac{dy}{1+y^2} \), involves a known integral resulting in \( \arctan(y) \).
  • The right side: \( \int x^2 dx \), results after calculation in \( \frac{x^3}{3} + C \), where \(C\) is the constant of integration.
After integrating both sides, they equal to \( \arctan(y) = \frac{x^3}{3} + C \). The integration helps us transition from a relationship in terms of derivatives to a function that describes \(y\) in terms of \(x\). Remember, it is important always to include the constant \(C\) as it represents an infinite number of possible solutions.
Family of Solutions
The family of solutions in a differential equation context refers to the infinite set of solutions that satisfy the differential equation. This family arises due to the constant of integration, \(C\), introduced during the integration process.
Each unique value of \(C\) corresponds to a different solution, representing a particular function that satisfies the original equation.
  • From our integrated equation: \( \arctan(y) = \frac{x^3}{3} + C \), solving for \(y\) reveals: \( y = \tan\left( \frac{x^3}{3} + C \right) \).
  • This solution is explicit because it directly expresses \(y\) in terms of \(x\), with \(C\) being any real number, showcasing an entire set of potential solutions.
Different values of \(C\) will adjust the angle inside the tangent function, creating a progression of curves that are all valid solutions to the equation. Thus, the concept of a 'family of solutions' is crucial as it underlines that the solution to differential equations is not just a single function, but a collection of them, influenced by initial conditions or constraints.

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Most popular questions from this chapter

A rocket, fired upward from rest at time \(t=0 .\) has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t .\) It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation$$ m \frac{d v}{d t}=c k-m g $$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t\) (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. \([\)Take \(g=9.8\) \(\left.\mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)

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Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) \(y^{\prime}=-x y\)

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