Question

Find a particular solution satisfying the given conditions. \(y y^{\prime \prime}+y^{\prime 2}+4=0 \quad y=3, y^{\prime}=0\) when \(x=1\)

Short answer

The particular solution is \( y = -\frac{2}{3} x^2 + \frac{4}{3} x + \frac{7}{3} \).

Step-by-step solution

- Understanding the Differential Equation

The given differential equation is: \[ y y^{\texttt{''}} + (y^{\texttt{'}})^2 + 4 = 0 \]We are also given the initial conditions: \[ y(1) = 3, \, y^{\texttt{'}}(1) = 0 \]

- Substituting Initial Conditions

Using the initial conditions, substitute \(y = 3\) and \(y^{\texttt{'}} = 0\) at \(x = 1\) into the differential equation: \[ 3 y^{\texttt{''}} + (0)^2 + 4 = 0 \]This simplifies to: \[ 3 y^{\texttt{''}} + 4 = 0 \]

- Solving for the Second Derivative

Solve for \(y^{\texttt{''}}\): \[ 3 y^{\texttt{''}} + 4 = 0 \]\[ 3 y^{\texttt{''}} = -4 \]\[ y^{\texttt{''}} = -\frac{4}{3} \]

- Finding Particular Solution

To find a particular solution, integrate the second derivative twice. First integration of \(y^{\texttt{''}} = -\frac{4}{3}\): \[ y^{\texttt{'}} = -\frac{4}{3} x + C_1 \]Using the initial condition \(y^{\texttt{'}} = 0\) when \(x = 1\): \[ 0 = -\frac{4}{3} (1) + C_1 \]\[ C_1 = \frac{4}{3} \]

- Second Integration

Integrate again to find \(y\): \[ y = -\frac{4}{3} \frac{x^2}{2} + C_1 x + C_2 \]\[ y = -\frac{2}{3} x^2 + \frac{4}{3} x + C_2 \]Using the initial condition \( y = 3 \) when \( x = 1 \): \[ 3 = -\frac{2}{3}(1)^2 + \frac{4}{3}(1) + C_2 \]\[ 3 = -\frac{2}{3} + \frac{4}{3} + C_2 \]\[ 3 = \frac{2}{3} + C_2 \]\[ C_2 = 3 - \frac{2}{3} = \frac{7}{3} \]

- Final Particular Solution

Putting it all together, the particular solution is: \[ y = -\frac{2}{3} x^2 + \frac{4}{3} x + \frac{7}{3} \]

Key concepts

Second-Order Differential Equations

A second-order differential equation involves the second derivative of a function. In simpler terms, it includes the rate of change of the rate of change of a function. In our problem, the differential equation is: \[ y y^{\texttt{''}} + (y^{\texttt{'}})^2 + 4 = 0 \]Here, \( y^{\texttt{''}} \) represents the second derivative of the function \( y \). The first step is to understand what each term implies. For example, \( y y^{\texttt{''}} \) tells us about the nonlinear nature of the equation because it involves the product of the function and its second derivative. On the other hand, \( (y^{\texttt{'}})^2 \) signifies the square of the first derivative. Such equations can describe many real-world phenomena like vibrations, heat conduction, and more.

Initial Conditions

Initial conditions are values given at specific points to help solve differential equations. They allow us to find particular solutions, rather than the general form. In our case, the given initial conditions are: \[ y(1) = 3, \, y^{\texttt{'}}(1) = 0 \]This means that when \( x = 1 \), \( y \) is 3 and the first derivative of \( y \) (\( y^{\texttt{'}} \)) is zero. These conditions are crucial because they allow us to find specific values for the constants arising during integration. By substituting these values into the differential equation, we can simplify the problem and progress toward finding a specific solution that fits the given scenario.

Integration Techniques

Integration is the process of finding the antiderivative, and it's essential for solving differential equations. Let's break down the steps:
First, we determined the second derivative: \[ y^{\texttt{''}} = -\frac{4}{3} \]Next, we integrated this result to find the first derivative: \[ y^{\texttt{'}} = -\frac{4}{3} x + C_1 \]Using the initial condition \( y^{\texttt{'}}(1) = 0 \), we found \( C_1 \): \[ C_1 = \frac{4}{3} \]Then, we integrated again to find \( y \): \[ y = -\frac{2}{3} x^2 + \frac{4}{3} x + C_2 \]By using the initial condition \( y(1) = 3 \), we solved for \( C_2 \): \[ C_2 = \frac{7}{3} \]Each step involves careful integration and using initial values to find constants. This process ensures that every solution fits the specific conditions provided.

Particular Solutions

A particular solution satisfies both the differential equation and given initial conditions. The general form of a solution involves arbitrary constants, but by plugging in the initial conditions, we solve for these constants. For instance, after integrating the given differential equation twice and applying our initial conditions, we arrived at: \[ y = -\frac{2}{3} x^2 + \frac{4}{3} x + \frac{7}{3} \]This is the particular solution that meets the original problem's requirements. Particular solutions are important in real-world applications because they provide precise answers adjusted for specific situations, making differential equations extremely useful in engineering, physics, and other fields. Once you understand the process, these solutions reveal the unique behavior of systems under specific conditions.