Question

Convert the following higher-order equations to systems of first-order equations: (a) \(Y^{\prime \prime \prime}+4 Y^{\prime \prime}+5 Y^{\prime}+2 Y=2 x^{2}+10 x+8\), \(Y(0)=1, \quad Y^{\prime}(0)=-1, \quad Y^{\prime \prime}(0)=3\) The true solution is \(Y(x)=e^{-x}+x^{2}\). (b) \(Y^{\prime \prime}+4 Y^{\prime}+13 Y=40 \cos (x), \quad Y(0)=3, \quad Y^{\prime}(0)=4\) The true solution is \(Y(x)=3 \cos (x)+\sin (x)+e^{-2 x} \sin (3 x)\).

Short answer

(a) \[\begin{cases} y_1' = y_2 \ y_2' = y_3 \ y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1 \ \text{with } y_1(0) = 1, y_2(0) = -1, y_3(0) = 3 \ \end{cases}\](b) \[\begin{cases} y_1' = y_2 \ y_2' = 40\cos(x) - 4y_2 - 13y_1 \ \text{with } y_1(0) = 3, y_2(0) = 4 \ \end{cases}\]}

Step-by-step solution

- Define the new variables for part (a)

Let us define the new variables to convert the third-order differential equation to a system of first-order equations: \[y_1 = Y, \quad y_2 = Y', \quad y_3 = Y''\]

- Write the equations for part (a)

From the definitions, the following first-order differential equations are derived:\[y_1' = y_2\]\[y_2' = y_3\]The original differential equation can be written in terms of these new variables:\[y_3' + 4y_3 + 5y_2 + 2y_1 = 2x^2 + 10x + 8\]

- Write the system for part (a)

Now compile the first-order system:\[y_1' = y_2\]\[y_2' = y_3\]\[y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1\]Initial conditions:\[y_1(0) = 1, \quad y_2(0) = -1, \quad y_3(0) = 3\]

- Define new variables for part (b)

Similarly, define the new variables to convert the second-order differential equation to a first-order system:\[y_1 = Y, \quad y_2 = Y'\]

- Write the equations for part (b)

From the definitions, the following first-order differential equations are derived:\[y_1' = y_2\]The original differential equation can be written in terms of these new variables:\[y_2' + 4y_2 + 13y_1 = 40\cos(x)\]

- Write the system for part (b)

Compile the first-order system:\[y_1' = y_2\]\[y_2' = 40\cos(x) - 4y_2 - 13y_1\]Initial conditions:\[y_1(0) = 3, \quad y_2(0) = 4\]

Key concepts

first-order systems

Converting higher-order differential equations into first-order systems is a common technique in solving complex problems. This method involves breaking down a higher-order equation into multiple first-order equations.
An equation's order is determined by the highest derivative present. For example, a third-order equation involves up to third derivatives. By introducing new variables to represent these derivatives, you can simplify the problem.
Consider the given problem: we have higher-order differential equations of orders three and two, respectively. To transform them into first-order systems, follow these steps:

• Introduce new variables for each derivative. For example, for the equation with a third derivative, define:
  \text{\(y_1 = Y\)}, \(y_2 = Y'\), and \(y_3 = Y''\).
• Rewrite the higher-order equation using these new variables.
• Establish first-order differential equations from the definitions and the original equation.
After this process, you will have transformed the problem into a more manageable system of first-order differential equations.

initial conditions

Initial conditions are critical when solving differential equations. They provide the specific values of the function and its derivatives at a given point, usually when \(x = 0\).
In this exercise, initial conditions for part (a) are \(Y(0) = 1\), \(Y'(0) = -1\), and \(Y''(0) = 3\). These conditions ensure a unique solution. For part (b), the initial conditions are \(Y(0) = 3\) and \(Y'(0) = 4\).
These initial conditions translate directly to the values of the newly defined variables at \(x = 0\). For example:

• For part (a): \(y_1(0) = 1\), \(y_2(0) = -1\), and \(y_3(0) = 3\).
• For part (b): \(y_1(0) = 3\) and \(y_2(0) = 4\).

By incorporating initial conditions, you can solve the first-order systems accurately. They help in determining the constants of integration and ensuring the solution matches the problem's specifics.

step-by-step solutions

Step-by-step solutions are essential for understanding complex mathematical problems. Breaking down the problem into manageable steps makes it easier to follow and solve efficiently.
Consider the solution process for the given exercise:

• Step 1: Define new variables for each derivative to simplify the higher-order equation into a first-order system. For part (a), we define: \(y_1 = Y\), \(y_2 = Y'\), and \(y_3 = Y''\).
• Step 2: Use these variables to rewrite the original higher-order differential equation in terms of the new variables. For example, \(Y^{\text{'''}} + 4Y'' + 5Y' + 2Y = 2x^2 + 10x + 8\) becomes \(y_3' + 4y_3 + 5y_2 + 2y_1 = 2x^2 + 10x + 8\).
• Step 3: Write the resulting system of first-order equations. For instance, \(y_1' = y_2\), \(y_2' = y_3\), and \(y_3' = 2x^2 + 10x + 8 - 4y_3 - 5y_2 - 2y_1\).
• Step 4: Apply the initial conditions to these new variables.

Following these steps ensures a clear and structured approach to solving differential equations.

differential equations

Differential equations involve functions and their derivatives. They are fundamental in modeling numerous phenomena in science and engineering.
In the given exercise, we deal with higher-order differential equations. For example, \(Y^{\text{'''}} + 4Y'' + 5Y' + 2Y = 2x^2 + 10x + 8\) and \(Y'' + 4Y' + 13Y = 40\text{cos}(x)\). These equations describe how a function and its derivatives relate to each other.
To solve these equations, especially when dealing with higher-order ones, converting them into a system of first-order differential equations is beneficial. It simplifies the process and makes numerical solutions more feasible.
Understanding the basic principles of differential equations is crucial. They help in predicting the behavior of various systems, from mechanical vibrations to electrical circuits. By learning to convert and solve these equations, one can unlock a deeper understanding of dynamic systems and their behaviors.