Problem 7 - Advanced Engineering Mathematics

Question

In Problems \(7-10, x=c_{1} \cos t+c_{2} \sin t\) is a two-parameter family of solutions of the second-order DE \(x^{\prime \prime}+x=0\). Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. $$ x(0)=-1, \quad x^{\prime}(0)=8 $$

Step-by-Step Solution

Answer

The solution is \( x = -\cos t + 8\sin t \).

Step 1: Identify the General Solution

The given two-parameter family of solutions is \( x = c_1 \cos t + c_2 \sin t \). This is the general solution to the differential equation \( x'' + x = 0 \), where the complementary solutions \( \cos t \) and \( \sin t \) are used.

Step 2: Apply the Initial Condition for x(t)

The initial condition given is \( x(0) = -1 \). Substitute \( t = 0 \) into the general solution:\[ x(0) = c_1 \cos(0) + c_2 \sin(0) = c_1 \cdot 1 + c_2 \cdot 0 = c_1. \]Therefore, \( c_1 = -1 \).

Step 3: Differentiate to Find x'(t)

Differentiate the general solution \( x = c_1 \cos t + c_2 \sin t \) with respect to \( t \) to find \( x'(t) \):\[ x'(t) = -c_1 \sin t + c_2 \cos t. \]

Step 4: Apply the Initial Condition for x'(t)

Use the initial condition \( x'(0) = 8 \). Substitute \( t = 0 \) into the expression for \( x'(t) \):\[ x'(0) = -c_1 \sin(0) + c_2 \cos(0) = -c_1 \cdot 0 + c_2 \cdot 1 = c_2. \]Thus, \( c_2 = 8 \).

Step 5: Write the Specific Solution

Now, substitute the values of \( c_1 \) and \( c_2 \) into the general solution:\[ x = -1 \cdot \cos t + 8 \cdot \sin t = -\cos t + 8\sin t. \]This is the solution to the initial value problem.

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with specific conditions given at a certain point, usually to determine which solution of the differential equation is sought. In this case, the initial value conditions given are \( x(0) = -1 \) and \( x'(0) = 8 \).
These conditions allow us to find specific solutions within the general family of solutions of a differential equation. By substituting the initial conditions into the general solution, which is typically an expression containing arbitrary constants, we can solve for these constants. This action transforms the general solution into a specific solution that fits the problem's initial conditions.

• The initial condition \( x(0) = -1 \) allows us to determine the constant \( c_1 \).
• The condition \( x'(0) = 8 \) helps find the value for \( c_2 \).

This step-by-step approach outlines how initial conditions tailor the general solution to meet specific criteria of the problem at hand.

Cosine and Sine Functions

The cosine and sine functions play a crucial role in solving second-order linear differential equations with constant coefficients. They are part of the set of solutions because they satisfy the homogeneous differential equation \( x'' + x = 0 \). If you differentiate the cosine and sine functions, you will notice they cycle back to their respective forms:
\(\cos'(t) = -\sin(t) \quad \text{and} \quad \sin'(t) = \cos(t)\)

• Due to these properties, cosine and sine represent the solutions to the characteristic equation of the differential equation.
• The general solution can usually be expressed as a linear combination of cosine and sine functions: \( c_1 \cos t + c_2 \sin t \).

These functions are fundamental in oscillatory systems and any problem involving harmonic motion, demonstrating their extensive applications in physics and engineering.

Differential Equations Solutions

Solving differential equations, particularly second-order ones, often involves finding a function that satisfies the differential equation throughout a given domain. For the equation \( x'' + x = 0 \), we begin by identifying solutions to the associated homogeneous problem, which are typically of exponential, sine, or cosine form.
In our example, the homogeneous solution encompasses the cosine and sine functions: \( x = c_1 \cos t + c_2 \sin t \). The task then becomes applying the initial conditions to pinpoint a particular solution from this general form.

• Start by substituting the initial conditions into the general solution, setting \( t = 0 \).
• Determine the values of \( c_1 \) and \( c_2 \) as shown through integration or initial condition application.

This structured method leads us to a specific solution that meets both the differential equation and the initial conditions provided, offering a complete and precise answer to the problem. Successfully solving such equations yields valuable insights into the behavior of dynamic systems, from mechanical vibrations to electrical circuits.