Question

$$ y^{(4)}-16 y=0, \quad y(0)=2, y^{\prime}(0)=2, y^{\prime \prime}(0)=-2, y^{\prime \prime \prime}(0)=0 $$$$ \begin{array}{l} y^{(4)}-6 y^{\prime \prime \prime}+7 y^{\prime \prime}+6 y^{\prime}-8 y=0, \quad y(0)=-2, \quad y^{\prime}(0)=-8, \quad y^{\prime \prime}(0)=-14, \\ y^{\prime \prime \prime}(0)=-62 \end{array} $$

Short answer

Answer: The particular solution for the given equation and initial conditions is y(x) = e^{2x} + e^{-2x} + \cos(2x).

Step-by-step solution

- Find General Solution for y^{(4)} - 16y = 0

To find the general solution, we need to find the characteristic equation: r^4 - 16 = 0. Solve for r's roots: r = ±2, ±2i.

- Write down the general solution

Using the roots from Step 1, the general solution for the first problem can be written as: y(x) = c_1 e^{2x} + c_2 e^{-2x} + c_3 \cos(2x) + c_4 \sin(2x)

- Apply the initial conditions

Use the given initial conditions: 1. y(0) = 2 ⇒ c_1 + c_2 = 2 2. y'(0) = 2 ⇒ 2c_1 - 2c_2 + 2c_4 = 2 3. y''(0) = -2 ⇒ 4c_1 + 4c_2 - 4c_3 = -2 4. y'''(0) = 0 ⇒ -8c_1 + 8c_2 - 8c_4 = 0

- Solve the system of equations for constants

Solve the system of linear equations from Step 3 to find the values of c_1, c_2, c_3, and c_4. After solving we get c_1 = 1, c_2 = 1, c_3 = 1, and c_4 = 0.

- Write down the particular solution

Insert the values of the constants into the general solution: y(x) = e^{2x} + e^{-2x} + \cos(2x) ##Problem 2

- Find the General Solution for y^{(4)} - 6y''' + 7y'' + 6y' - 8y = 0

This is already a simpler equation with no need to find a characteristic equation. The general solution is: y(x) = (Ax^3 + Bx^2 + Cx + D) e^{2x}

- Apply the initial conditions

Use the given initial conditions: 1. y(0)=-2 ⇒ D=-2 2. y'(0)=-8 ⇒ -10A + 12B - 12C - 8 = -8 3. y''(0)=-14 ⇒ 44A - 24B + 20C - 14 =-14 4. y'''(0)=-62 ⇒ -164A + 96B - 38C - 62 = -62

- Solve the system of equations for constants

Solve the system of linear equations from Step 2 to find the values of A, B, and C. After solving we get A = 2, B = 3, and C = -2.

- Write down the particular solution

Insert the values of the constants into the general solution: y(x) = (2x^3 + 3x^2 - 2x - 2) e^{2x}

Key concepts

Characteristic Equation

Understanding the characteristic equation is essential when solving linear homogenous differential equations. This equation is derived by substituting a solution of the form \(e^{rx}\) into the homogeneous differential equation, resulting in a polynomial equation in terms of \(r\).

For the given exercise, the characteristic equation for the first differential equation \(y^{(4)} - 16y = 0\) is \(r^4 - 16 = 0\). By solving for \(r\), we determine the roots that are the backbone for constructing the general solution. These roots represent the different behaviors of the solution components such as exponential growth, decay, and oscillations.

Initial Conditions

Initial conditions are crucial in determining the particular solution to a differential equation. They are the values of the function and its derivatives at a specific point, usually \(x = 0\).

In our examples, we use the given initial conditions to find the constants in the general solutions. For instance, in the first equation, we used the conditions \(y(0) = 2\), \(y'(0) = 2\), \(y''(0) = -2\), and \(y'''(0) = 0\) to find the constants \(c_1\), \(c_2\), \(c_3\), and \(c_4\). Applying initial conditions links the abstract general solution to a specific scenario, making the solution applicable and meaningful.

General Solution

The general solution to a differential equation involves all possible solutions and contains arbitrary constants that can be determined using initial conditions or boundary conditions. It encompasses all behaviours predicted by the characteristic equation.

For the first differential equation in our exercise, the general solution is expressed as \(y(x) = c_1 e^{2x} + c_2 e^{-2x} + c_3 \cos(2x) + c_4 \sin(2x)\), which combines exponential and trigonometric functions, indicating a mix of growing, decaying, and oscillating solutions. Such a form allows the solution to be tailored to specific needs by setting the appropriate constants.

Boundary Value Problems

Boundary value problems involve differential equations with conditions specified at different points, often at the boundaries of the domain of interest. While the exercise provided doesn't present a classical boundary value problem (since it offers initial conditions instead), understanding boundary value problems is still worthwhile.

These problems are commonly encountered in physics and engineering and may require specific techniques for solution. Such conditions may specify the value of the solution at two or more points, determining a unique solution to what would otherwise be a general solution with multiple possibilities.