Chapter 8: Problem 28
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
Expert verified
The second derivative is \( y^{\backprime\backprime} = -\frac{4}{3} \).
Step by step solution
01
Identify the given conditions
The given differential equation is \[ y y^{\backprime\backprime} + y^{\backprime 2} + 4 = 0 \]The initial conditions are \[ y(1) = 3 \] and \[ y^{\backprime}(1) = 0 \].
02
Substitute initial conditions into the differential equation
Substitute \( y = 3 \) and \( y^{\backprime} = 0 \) into the equation: \[ 3 \times y^{\backprime\backprime} + 0^2 + 4 = 0 \] It simplifies to \[ 3 y^{\backprime\backprime} + 4 = 0 \].
03
Solve for \( y^{\backprime\backprime} \)
Rearrange the simplified equation to solve for the second derivative \( y^{\backprime\backprime} \): \[ 3 y^{\backprime\backprime} = -4 \]\[ y^{\backprime\backprime} = -\frac{4}{3} \].
04
Verify the particular solution
Check that the second derivative found satisfies the original differential equation with the given conditions. Since all initial conditions are satisfied, the solution for \( y^{\backprime\backprime} \) holds true in this equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
initial conditions
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:
- \(y(1) = 3\)
- \(y'(1) = 0\)
second derivative
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.
\[ y y^{\backprime\backprime} + y^{\backprime 2} + 4 = 0 \]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.
particular solution
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.
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
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 \(y^{\backprime\backprime} = -\frac{4}{3}\).