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Chapter 10: Problem 7

\(y^{\prime}+y^{2}=0\).

Short Answer

Expert verified
The solution is \( y = \frac{1}{x + C} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is \( y' + y^2 = 0 \). This is a first-order ordinary differential equation and is separable, meaning it can be rewritten as separate functions of \( y \) and \( x \).
02

Rearrange the Equation for Separation

Rearrange the equation to explicitly separate the variables: \( y' = -y^2 \). This allows us to separate \( y \) on one side and \( x \) on the other.
03

Separate Variables

Rewrite \( y' = \frac{dy}{dx} \) and separate the variables: \( \frac{dy}{y^2} = -dx \).
04

Integrate Both Sides

Integrate both sides of the separated equation: \( \int \frac{dy}{y^2} = \int -dx \). This gives \( -\frac{1}{y} = -x + C_1 \), where \( C_1 \) is the constant of integration.
05

Solve for y

Rearrange the equation to solve for \( y \): \( \frac{1}{y} = x + C_1 \), therefore, \( y = \frac{1}{x + C_1} \). This is the general solution of 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 Ordinary Differential Equations
First-order ordinary differential equations are fundamental in calculus and mathematics. These equations involve derivatives of a function with respect to one variable and are of the first order since they only contain first derivatives. In the equation given, \( y'+y^2=0 \), \( y' \) represents the derivative of \( y \) with respect to \( x \).
Understanding these types of equations is crucial, as they often model important real-world phenomena like population growth, heat transfer, and financial systems.
  • These equations are represented in the format \( \, f(x,y,y') = 0 \, \), focusing on the relationship between the function and its derivative.
  • The main challenge is to find the function \( y \, \) that satisfies such an equation.
  • These equations could be linear or nonlinear, as in our example, because of the \( y^2 \, \) term.
Variable Separation
A powerful technique for solving first-order differential equations is variable separation. Here, you rewrite the equation in a way that allows you to separate and relocate variables on different sides of the equation.
In the example equation, \( y' + y^2 = 0 \), the goal is to express it so that all \( y \, \) terms are on one side and all \( x \, \) terms are on the other. We achieve this by:
  • Rewriting \( y' \) as \( \frac{dy}{dx} \) to better expose the differential nature of the terms.
  • Then, rearranging to yield \( \frac{dy}{y^2} = -dx \), making it easier to integrate both sides separately.
This separation is important because it sets the stage for integration, enabling you to integrate both sides with respect to their own variables. Simplifying the problem allows solving for \( y \, \) as a function of \( x \).
Integration Techniques
Once variables are separated, integration becomes the tool to solve for the function. Integration involves finding a function whose derivative matches given terms. For the equation from the example, \( \frac{dy}{y^2} = -dx \), you need to integrate each side.
  • The left side, \( \int \frac{dy}{y^2} \), uses the power rule, resulting in the integration of \( y^{-2} \, \).
  • The right side, \( \int -dx \), is elementary and results in \( -x \, \).
Through integration, you attain the equation \( -\frac{1}{y} = -x + C_1 \), where \( C_1 \) is an arbitrary constant. This constant captures all potential solutions, highlighting the indefinite nature of integration unless further initial conditions are given.
General Solution of Differential Equations
The ultimate objective when solving differential equations is obtaining the general solution. This solution encapsulates all possible specific solutions of a differential equation using arbitrary constants.
From our integrated equation, \( -\frac{1}{y} = -x + C_1 \), rearranging terms helps isolate \( y \). By doing algebraic manipulation, we find that \( y = \frac{1}{x + C_1} \).

  • The general solution shows how \( y \, \) depends on \( x \, \) and includes the constant \( C_1 \), signifying any initial condition can determine a particular solution.
  • This constant stems from the indefinite integral and represents an integration constant unique to each problem setup.
  • General solutions offer the flexibility needed to adapt the solution to initial values or boundary conditions in real-life scenarios.
Understanding that these constants provide a family of solutions helps in modeling diverse situations by tweaking the initial condition related to \( C_1 \).

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Most popular questions from this chapter

In Exercises 12 - 17, determine the independent variable, and use dsolve to find the solution to the indicated initial value problem. Use ezplot to plot the solution over the indicated time interval. \(y^{\prime}+t y=t, y(0)=-1,[-4,4]\).

\(y^{\prime}+y^{2}=0, y(0)=2,[0,5]\).

In Exercises 18 and 19, use dsolve to obtain the solution of each of the indicated second order differential equations. Use the simple command to find the simplest form of that solution. Use ezplot to sketch the solution on the indicated time interval. $$ \begin{aligned} &y^{\prime \prime}+4 y=3 \cos 2.1 t \\ &y(0)=0, y^{\prime}(0)=0,[0,64 \pi] \end{aligned} $$

Use dsolve to solve the initial value problem $$ \begin{aligned} &y^{\prime}=v \\ &v^{\prime}=-2 v-2 y+\cos 2 t \end{aligned} $$ with \(y(0)=1\) and \(v(0)=-1\), over the time interval \([0,30]\).

In Exercises 18 and 19, use dsolve to obtain the solution of each of the indicated second order differential equations. Use the simple command to find the simplest form of that solution. Use ezplot to sketch the solution on the indicated time interval. $$ \begin{aligned} &y^{\prime \prime}+16 y=3 \sin 4 t \\ &y(0)=0, y^{\prime}(0)=0,[0,32 \pi] \end{aligned} $$
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