Question

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \(y^{\prime}+2 x y^{2}=0, \quad y=1\) when \(x=2\)

Short answer

The particular solution is \( y = \frac{1}{x^2 - 3} \) with the boundary condition y(2) = 1.

Step-by-step solution

- Rewrite the Differential Equation

Given the differential equation: \( y' + 2xy^2 = 0 \)First, rewrite it in a form suitable for separating variables: \( y' = -2xy^2 \).

- Separate Variables

Separate the variables to place all terms involving 'y' on one side and all terms involving 'x' on the other side:\( \frac{dy}{y^2} = -2x \, dx \).

- Integrate Both Sides

Integrate both sides of the equation:\( \int \frac{dy}{y^2} = \int -2x \, dx \).The left side integrates to \( -\frac{1}{y} \) and the right side to \( -x^2 + C \), where C is the integration constant:\( -\frac{1}{y} = -x^2 + C \).

- Solve for y

Rewrite the equation in terms of 'y':\( \frac{1}{y} = x^2 - C \).Invert both sides to solve for 'y':\( y = \frac{1}{x^2 - C} \).

- Apply the Boundary Condition to Find C

Use the boundary condition y(2) = 1 to find the constant C:\( 1 = \frac{1}{2^2 - C} \).Solving for C, we get:\( 2^2 - C = 1 \).\( 4 - C = 1 \).\( C = 3 \).

- Write the Particular Solution

Substitute the value of C back into the general solution:\( y = \frac{1}{x^2 - 3} \).

- Plot the Slope Field and Solution Curves

Use a computer software or graphing tool to plot the slope field of the differential equation and the solution curves. This visual representation helps to understand the behavior of solutions for different initial conditions.

Key concepts

Separation of Variables

In the context of differential equations, separation of variables is a technique to simplify and solve equations by separating the variables on different sides of the equation. Start by rewriting the equation in a form where all terms involving one variable (e.g., y) are on one side, and all terms involving the other variable (e.g., x) are on the other side. For instance, let’s consider the differential equation provided: \( y' + 2xy^2 = 0 \). We first rewrite this equation to isolate \( y' \): \( y' = -2xy^2 \). Now, we separate the variables: \( \frac{dy}{y^2} = -2x \, dx \). This step is crucial for simplifying the problem to allow direct integration, where both sides of the equation can be integrated independently. Think of it like sorting out different items into separate boxes based on their characteristics.

Boundary Conditions

Boundary conditions are the specific values that a solution to a differential equation must satisfy. They are used to determine the constants of integration that appear after solving the equation. Essentially, these conditions 'anchor' the solution to a specific point. In the example provided, the boundary condition is \( y = 1 \) when \( x = 2 \). After separating the variables and integrating to find a general solution, we apply this condition to find the particular solution. With our general solution as \( y = \frac{1}{x^2 - C} \), we substitute the given boundary values: \( 1 = \frac{1}{2^2 - C} \). Solving this equation helps us determine the constant \(C\): \( 2^2 - C = 1 \). Simplifying, we get \(C = 3\). This particularization is essential to make the solution unique and applicable to real-world problems where such conditions are often present.

Integration

Integration is a core mathematical operation used to solve continuous functions, and it plays a vital role in solving differential equations. After separating the variables, integrate both sides of the equation. Using our example, after separation, we have \( \frac{dy}{y^2} = -2x \, dx \). Now, integrate each side: \( \int \frac{dy}{y^2} = \int -2x \, dx \). The left side integrates to \(-\frac{1}{y}\) and the right side integrates to \( -x^2 + C \), where \( C \) is the integration constant: \( -\frac{1}{y} = -x^2 + C \). Integration transforms a differential equation into a solvable form. The constants of integration need to be determined using boundary conditions to provide precise solutions. This technique is pivotal in transforming complex, variable-dependent problems into manageable equations.