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

Initial conditions are specific values given for the function and its derivatives at a particular point. They help in finding a unique solution to a differential equation. For our exercise, the initial conditions are given as follows:

• \(y(1) = 3\)
• \(y'(1) = 0\)

These values provide the necessary data to integrate the differential equation and find a particular solution. By plugging these into the differential equation, we can simplify our problem considerably. Initial conditions are essential because differential equations can often have multiple solutions. These conditions help in pinpointing the exact solution that fits the problem requirements.

In differential equations, the second derivative \(y''\) represents the rate at which the first derivative \(y'\) is changing. It's a key term in understanding the behavior of the function. In our exercise, the differential equation given is:
\[ y y^{\backprime\backprime} + y^{\backprime 2} + 4 = 0 \]By substituting the initial conditions \(y(1) = 3\) and \(y'(1) = 0\), we can reframe this equation to:
\[3 \times y^{\backprime\backprime} + 0^2 + 4 = 0\] Which further simplifies to:
\[3 y^{\backprime\backprime} + 4 = 0\]To isolate \(y^{\backprime\backprime}\), we rearrange the equation:
\[3 y^{\backprime\backprime} = -4\]\[ y^{\backprime\backprime} = -\frac{4}{3}\]This tells us the value of the second derivative under the given conditions, which is crucial for understanding the curvature and concavity of the function.

A particular solution is a specific solution to a differential equation that satisfies the initial conditions given in the problem. While a general solution involves arbitrary constants and represents a family of possible solutions, a particular solution is unique and tailored to specific initial values.
In this exercise, the particular solution is derived by solving the differential equation with the initial conditions \(y(1) = 3\) and \(y'(1) = 0\). By substituting these initial conditions into the differential equation, we eliminated arbitrary constants and directly solved for the specific value:
\[y^{\backprime\backprime} = -\frac{4}{3}\]Thus, our particular solution is verified to satisfy both the differential equation and the initial conditions, making it the unique solution to this problem.

Simplifying differential equations involves making the equation easier to solve by utilizing given conditions and algebraic manipulations. Here’s how we did it in this exercise:

• Firstly, we identified and used the initial conditions \(y(1) = 3\) and \(y'(1) = 0\).
• By substituting these values into the differential equation \(yy^{\backprime\backprime} + y^{\backprime 2} + 4 = 0\), the equation simplified to \(3y^{\backprime\backprime} + 4 = 0\).
• Finally, we isolated the second derivative term \(y^{\backprime\backprime}\), leading to \(y^{\backprime\backprime} = -\frac{4}{3}\).

These steps illustrate the process of reducing complexity in a problem. Often, simpler forms reveal the solution more clearly. Simplifying differential equations can involve various techniques, such as substitution, separation of variables, and integrating factors.