Question

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime}+2 x y^{2}=0\) \(y=1\) when \(x=2\)

Short answer

The general solution is \( y = \frac{1}{x^2 - C} \). The particular solution is \( y = \frac{1}{x^2 - 3} \).

Step-by-step solution

- Rearrange the Differential Equation

Start by rearranging the given differential equation \( y' + 2xy^2 = 0 \) so that the derivatives are on one side and the other terms on the other side. This gives \( y' = -2xy^2 \).

- Separate the Variables

Separate the variables to enable integration. This means writing it as \( \frac{dy}{y^2} = -2x dx \).

- Integrate Both Sides

Integrate both sides of the equation. Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[ \int \frac{1}{y^2} \, dy = \int -2x \, dx \].

- Solve the Integrals

Solve the integrals separately. The left side integral is \( \int y^{-2} \, dy = -y^{-1} \). The right side integral is \( \int -2x \, dx = -x^2 \). This leads to \( -\frac{1}{y} = -x^2 + C \), where \( C \) is the integration constant.

- Solve for y

Rearrange the equation to solve for \( y \): \( -\frac{1}{y} = -x^2 + C \) becomes \( \frac{1}{y} = x^2 - C \). Then \( y = \frac{1}{x^2 - C} \)

- Apply Boundary Condition

Use the boundary condition \( y = 1 \) when \( x = 2 \) to find the particular solution. Substitute these values into the general solution: \( 1 = \frac{1}{4-C} \). Solving for \( C \) gives \( C = 3 \).

- Write the Particular Solution

Substitute \( C = 3 \) back into the general solution \( y = \frac{1}{x^2 - C} \). The particular solution is \( y = \frac{1}{x^2 - 3} \).

Key concepts

Separation of Variables

Separation of variables is a fundamental method used to solve first-order differential equations. It's especially useful when we can arrange the equation such that each variable and its differential are on opposite sides. This way, each side of the equation contains only one variable, which simplifies the problem significantly.
For instance, consider the differential equation in our original exercise:
\text{Step 1: Begin by rearranging the given differential equation:} \( y' + 2xy^2 = 0 \).
Rearrange to isolate the derivative: \( y' = -2xy^2 \).
\text{Step 2: Separate the variables to enable integration:} \( \frac{dy}{y^2} = -2x dx \).
This step makes it clear how separation of variables works. It allows us to prepare each part for integration, making the equation easier to handle.

Boundary Conditions

Boundary conditions are specific values that a solution to a differential equation must satisfy. These conditions often help in finding a particular solution to a differential equation. In the given exercise, we've found the general solution first, which involved an arbitrary constant, and then applied the boundary conditions to find a specific value for that constant.
In our exercise, we had:

• General solution: \( y = \frac{1}{x^2 - C} \)
• Boundary condition: \( y = 1 \) when \( x = 2 \)

By substituting the boundary condition into the general solution, we get:
\( 1 = \frac{1}{4 - C} \).
Solving this, we find \( C = 3 \), giving us the particular solution: \( y = \frac{1}{x^2 - 3} \).
Boundary conditions are critical in differential equations because they allow us to hone in on a more precise solution that fits our problem's specific needs.

Integration

Integration is at the heart of solving differential equations, especially when using the separation of variables method. Once the equation is separated, each side is integrated with respect to its corresponding variable.
Let's revisit our exercise. After separating variables, we had:
\( \frac{dy}{y^2} = -2x dx \).
The next step is to integrate both sides:

• Left side: \( \int y^{-2} \, dy = -y^{-1} \)
• Right side: \( \int -2x \, dx = -x^2 \)

Combining these, we get: \( -\frac{1}{y} = -x^2 + C \).
Integration often introduces an integration constant, represented here by \( C \), which plays a crucial role in finding both the general and particular solutions.
Solving the integrals carefully while keeping the integration constants in mind is essential for the accuracy of the final solution.