Question

Each initial value problem was obtained from an initial value problem for a higher order scalar differential equation. What is the corresponding scalar initial value problem? $$ \mathbf{y}^{\prime}=\left[\begin{array}{c} y_{2} \\ y_{3} \\ y_{4} \\ y_{2}+y_{3} \sin \left(y_{1}\right)+y_{3}^{2} \end{array}\right], \quad \mathbf{y}(1)=\left[\begin{array}{r} 0 \\ 0 \\ -1 \\ 2 \end{array}\right] $$

Short answer

Question: Write down the corresponding scalar initial value problem for the given system of first-order differential equations with initial values y(1) = 0, y2(1) = 0, y3(1) = -1, and y4(1) = 2. Answer: The corresponding scalar initial value problem is: $$ \begin{cases} y_{1}^{\prime\prime\prime\prime} = y_{2} + y_{3}\sin(y_{1}) + y_{3}^{2} \\ y_{1}(1)=0 \\ y_{1}^\prime(1)= 0 \\ y_{1}^{\prime\prime}(1)= -1 \\ y_{1}^{\prime\prime\prime}(1)= 2 \end{cases} $$

Step-by-step solution

Write the given system of equations separately

We are given the following system of equations: $$ \begin{cases} y_{1}^\prime = y_{2} \\ y_{2}^\prime = y_{3} \\ y_{3}^\prime = y_{4} \\ y_{4}^\prime = y_{2} + y_{3}\sin(y_{1}) + y_{3}^{2} \end{cases} $$

Differentiate equations

To eliminate terms, we need to find higher order derivatives. First, we differentiate the \(y_1^\prime = y_2\). Then we will differentiate the obtained result and use the other equations to eliminate the terms as needed. $$ y_{1}^{\prime\prime} = y_{2}^{\prime} = y_{3} $$ Now, differentiate \(y_{1}^{\prime\prime} = y_{3}\), $$ y_{1}^{\prime\prime\prime} = y_{3}^{\prime} = y_{4} $$ Next, differentiate \(y_{1}^{\prime\prime\prime} = y_{4}\), $$ y_{1}^{\prime\prime\prime\prime} = y_{4}^{\prime} = y_{2} + y_{3}\sin(y_{1}) + y_{3}^{2} $$

Write the corresponding initial value problem

Now that we have found the fourth-order scalar differential equation, we can write down the corresponding initial value problem using the given initial values. From the given initial values, we can derive the required conditions: $$ \begin{cases} y_{1}(1)=0 \\ y_{1}^\prime(1)= y_2(1) = 0 \\ y_{1}^{\prime\prime}(1)= y_3(1) = -1 \\ y_{1}^{\prime\prime\prime}(1)= y_4(1) = 2 \end{cases} $$ So, the corresponding scalar initial value problem is: $$ \begin{cases} y_{1}^{\prime\prime\prime\prime} = y_{2} + y_{3}\sin(y_{1}) + y_{3}^{2} \\ y_{1}(1)=0 \\ y_{1}^\prime(1)= 0 \\ y_{1}^{\prime\prime}(1)= -1 \\ y_{1}^{\prime\prime\prime}(1)= 2 \end{cases} $$

Key concepts

Higher Order Differential Equations

Higher order differential equations involve derivatives of an order higher than one. For example, if we differentiate a function multiple times, we can end up with second, third, or even higher order derivatives. These equations often arise when the system or function being studied has complex dynamics.
Take for instance a fourth-order differential equation. This involves a fourth derivative, which relates to the rate at which the third-order rate changes. That's quite intricate, isn’t it? In this context, the understanding of how to deal with these complex relationships is essential.

• In practical applications, higher order differential equations might describe phenomena such as motion, where acceleration, or even change in acceleration, is essential to understand the behavior of a system.
• Solving these equations requires techniques such as differentiation and substitution, often alongside initial conditions to determine a unique solution.

System of Equations

When dealing with a system of equations, we often work with multiple equations that describe different aspects of a single situation. In the given exercise, we encounter several equations that are interdependent. By writing them separately, we can better understand their relationships.
Here's what to consider when approaching such problems:

• Each equation typically provides information about different variables or different states in the system.
• Solving these systems often involves eliminating certain variables or integrating certain parts to understand how they interconnect.

In this problem, we see how each equation is differentiated to form a more straightforward scalar equation representing the original system's dynamics. Bringing these together forms an important set of data points to understand behavior within a given framework.

Scalar Differential Equation

Scalar differential equations are often simpler forms arising from systems of equations. Essentially, they reduce complex systems into single equations involving one main scalar variable. The process involves expressing the system in terms of one leading variable and its derivatives.
This transformation is beneficial because:

• It simplifies the system by focusing on one primary variable, often making analysis and solutions more straightforward.
• Provides a clear path to solve higher order equations through methods like substitution, performing derivations, and using initial conditions.

In the presented problem, converting the system of equations into the scalar form helps distill the complex interrelations into one manageable equation, guiding us toward a clear solution approach.

Initial Conditions

Initial conditions are essential in solving differential equations because they provide the necessary framework to determine a unique solution. Without these, you may have infinitely many solutions.
For instance, in our given problem, having initial values at a specific point, like when \( t = 1 \), allows us to craft specific values of the function and its derivatives.
These are typically noted as:

• The value of the function at a particular point, \( y(1) = 0 \).
• Values of derivatives at that point, e.g., \( y' (1)\), \( y'' (1)\), \( y''' (1)\), etc.

These specific values serve as a pivot, enabling us to 'anchor' our equation and make it solveable, demystifying the complex interdependent systems into a lone, distinct result that suits the specific conditions at hand.