Question

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem. $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$

Short answer

Question: Write the implicit solution to the given initial value differential equation and explain how to determine which function corresponds to the initial value problem using graphing software. Answer: The implicit solution to the initial value differential equation is given by: $$-\frac{1}{4}\left( \frac{1}{2 + y^2} + \frac{1}{3} \right) = \int \frac{2x}{(2+y^{2})^{2}} dx$$ To determine which function corresponds to the initial value problem y(1)=-1, we can use graphing software to plot the solution and look for the function that passes through the given point (1, -1).

Step-by-step solution

Identify and rewrite the given differential equation

The given initial value differential equation is: $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$ Our goal is to solve this equation to find the function y(x).

Separate variables

To solve this differential equation, let's perform the separation of variables. We'll rewrite the equation as follows: $$\frac{y}{\left(2+y^2\right)^2} y^{\prime}(x)=\frac{2x}{(2+y^2)^2}$$ Now separate variables: $$\frac{y}{\left(2+y^2\right)^2} dy = \frac{2x}{(2+y^2)^2} dx$$

Integrate both sides

Now, we need to integrate both sides of the equation with respect to their respective variables: $$\int \frac{y}{(2+y^{2})^{2}} dy = \int \frac{2x}{(2+y^{2})^{2}} dx + C$$ The given integral on the left side does not have an elementary antiderivative, but we can rewrite it and evaluate it as follows: $$\int \frac{y}{(2+y^{2})^{2}} dy = -\frac{1}{4} \int \frac{d(2 + y^2)}{(2+y^2)^2} = -\frac{1}{4}\left( \frac{1}{2 + y^2} + C_1 \right)$$ On the right side, the integral depends on the unknown function y(x), so we leave it uncomputed and get the following implicit form: $$-\frac{1}{4}\left( \frac{1}{2 + y^2} + C_1 \right) = \int \frac{2x}{(2+y^{2})^{2}} dx + C$$ Now, apply the initial condition y(1)=-1: $$-\frac{1}{4}\left( \frac{1}{2 + (-1)^2} + C_1 \right) = \int_1^{-1} \frac{2x}{(2+y^{2})^{2}} dx$$ Plug in y(1)=-1 and find C_1: $$-\frac{1}{4}\left( \frac{1}{3} + C_1 \right) = 0 \Rightarrow C_1 = \frac{1}{3}$$

Write the implicit solution

Rewrite the implicit form of the solution with the constant C_1: $$-\frac{1}{4}\left( \frac{1}{2 + y^2} + \frac{1}{3} \right) = \int \frac{2x}{(2+y^{2})^{2}} dx$$ Now we have the implicit solution to the differential equation with the initial value problem. Use graphing software to plot the solution and determine which function corresponds to the solution of the given initial value problem.

Key concepts

Initial Value Problems

An initial value problem in differential equations involves finding a function that satisfies a given differential equation and also passes through a specified point known as the initial condition. Essentially, it is about solving a differential equation and specifying the particular solution that matches the initial condition at a certain point.
This process ensures that we get one unique solution among the infinitely many solutions that a differential equation can have. In this exercise, our initial value is given as \(y(1) = -1\), meaning at \(x = 1\), the value of the function \(y\) is \(-1\). This information is critical in determining the constant \(C_1\) necessary for our integration step.

• The initial condition provides a unique solution to the differential equation.
• This is achieved by determining the constant term arising from integration.
• The initial value is essential for precise solutions, especially in real-world applications where specific starting conditions are known.

Applying the initial condition helps tailor the general solution to fit particular scenarios integral to scientific computations.

Separation of Variables

Separation of variables is a method used to solve ordinary differential equations. It involves re-arranging the equation such that each variable appears on one side of the equation. This helps in making the equation easier to integrate.
In our differential equation \(y y^{\prime}(x)=\frac{2 x}{(2+y^{2})^{2}}\), we first rewrite it in a form where terms involving \(y\) and its derivative \(dy\) fall on one side and terms involving \(x\) and \(dx\) on the other. This results in:
\[ \frac{y}{(2+y^2)^2} dy = \frac{2x}{(2+y^2)^2} dx \]

• This technique simplifies the problem into an integration of two separate functions.
• It’s ideal for dealing with equations that can be factorized or rearranged to isolate variables.
• Separation of variables is straightforward for first-order ordinary differential equations and serves as a foundation for solving more complex problems.

Once separated, the next step is to integrate both sides to find the implicit solution, which we'll cover next.

Implicit Solution Integration

The integration of implicit solutions involves finding the integral of both sides of a separated differential equation. An implicit function is one where the solution is not immediately solved for a single variable but still defines the relationship between the variables involved.
In our task, integration was needed on both sides of the equation derived from the separation of variables. The left side was \( \int \frac{y}{(2+y^2)^2} dy \), which required creative techniques as it didn't directly result in a simple elementary function.
After applying integration techniques, we arrived at:
\[-\frac{1}{4}\left( \frac{1}{2 + y^2} + C_1 \right) = \int \frac{2x}{(2+y^{2})^{2}} dx \]

• Implicit solutions often require substitution or partial fraction decomposition.
• Integration of complex terms may involve special techniques beyond basic integration.
• The final answer presents the integral form that includes a constant involving the initial condition.

Using the implicit solution and confirming with the initial condition ensures the differential equation problems are resolved accurately, offering precise guidance in applications ranging from physics to engineering.