Problem 11 - Calculus: Early Transcendentals

Question

Solve the initial value problem \(y^{\prime \prime}+4 y=0, y(0)=\sqrt{3}, y^{\prime}(0)=2\).

Step-by-Step Solution

Answer

The solution is \(y(t) = \sqrt{3} \cos(2t) + \sin(2t)\).

Step 1: Identifying the Differential Equation Type

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients, expressed as \(y'' + 4y = 0\).

Step 2: Solving the Characteristic Equation

For the equation \(y'' + 4y = 0\), we assume a solution of the form \(y = e^{rt}\) and substitute this into the equation to get the characteristic equation \(r^2 + 4 = 0\).

Step 3: Finding the Roots of the Characteristic Equation

Solve the characteristic equation \(r^2 + 4 = 0\). This gives \(r^2 = -4\), leading to the roots \(r = \pm 2i\).

Step 4: Writing the General Solution

With the roots \(r = \pm 2i\), the general solution to the differential equation is \(y(t) = C_1 \cos(2t) + C_2 \sin(2t)\).

Step 5: Applying Initial Conditions to Find Constants

Use the initial conditions to find \(C_1\) and \(C_2\). First, set \(t = 0\): \(y(0) = C_1\). Given \(y(0) = \sqrt{3}\), we have \(C_1 = \sqrt{3}\).

Step 6: Differentiating the General Solution

Find \(y'(t)\) by differentiating the general solution: \(y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t)\).

Step 7: Using the Second Initial Condition

Use the condition \(y'(0) = 2\) to find \(C_2\). At \(t = 0\), \(y'(0) = 2C_2 = 2\), which implies \(C_2 = 1\).

Step 8: Writing the Particular Solution

Substitute \(C_1\) and \(C_2\) back into the general solution to get \(y(t) = \sqrt{3} \cos(2t) + \sin(2t)\). This is the particular solution to the initial value problem.

Key Concepts

Second-order Linear Homogeneous

A second-order linear homogeneous differential equation is a type of differential equation crucial in mathematical modeling, particularly in physics and engineering.
Such equations involve a second derivative of a function and can be written in the standard form:
\[ a y'' + b y' + c y = 0 \]where \(a\), \(b\), and \(c\) are constants. Important features of these equations include their linearity and lack of a non-zero term on the right-hand side.
This gives them the 'homogeneous' distinction, meaning the solution structure remains consistent, allowing the superposition of solutions. In this particular exercise, the equation is a simplified version: \(y'' + 4y = 0\), indicating constant coefficients with no first derivative term.

Characteristic Equation

The characteristic equation is a fundamental tool for solving second-order linear homogeneous differential equations with constant coefficients.
Assuming a solution in the exponential form \(y = e^{rt}\), substitute this into the homogeneous equation to derive the characteristic equation:

• \( y'' + 4y = 0 \)
• Substitute \(y = e^{rt}\): \(r^2 e^{rt} + 4e^{rt} = 0\)
• Thus, the characteristic equation becomes \( r^2 + 4 = 0 \)

In solving the equation \( r^2 + 4 = 0 \), we find complex roots \( r = \pm 2i \). These complex roots indicate that the differential equation will have solutions involving trigonometric functions, specifically cosine and sine.

Initial Value Problem

An initial value problem in differential equations refers to solving the equation with specific initial conditions provided.
Here, the initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) are used to find specific solutions for the constants in the general solution.

• Step 1: Express the general solution as \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \)
• Step 2: Apply \( y(0) = \sqrt{3} \) to find \( C_1 = \sqrt{3} \)
• Step 3: Differentiate the general solution to find \( y'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t) \)
• Step 4: Use \( y'(0) = 2 \) to establish \( C_2 = 1 \)

By solving these steps, we have defined the problem's constraints and found a solution specific to these initial values.

General Solution

The general solution of a differential equation represents a family of functions that includes all possible solutions to the equation.
Given the roots \( r = \pm 2i \), the general solution for the differential equation \( y'' + 4y = 0 \) is derived using complex exponentials:

• The solution takes the form \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \)
• This involves arbitrary constants \( C_1 \) and \( C_2 \), which can adapt to various initial conditions

This general solution reflects the characteristic behavior expected from systems modeled by differential equations with oscillatory solutions, often applicable to mechanical or electrical systems.

Particular Solution

The particular solution is a single specific solution derived from the general solution by applying initial conditions.
From our problem, after identifying the general solution \( y(t) = C_1 \cos(2t) + C_2 \sin(2t) \), initial conditions \( y(0) = \sqrt{3} \) and \( y'(0) = 2 \) were used to determine \( C_1 \) and \( C_2 \) as:

• \( C_1 = \sqrt{3} \)
• \( C_2 = 1 \)

Substituting these constants back into the general solution gives us the particular solution:
\[ y(t) = \sqrt{3} \cos(2t) + \sin(2t) \]This solution satisfies both the differential equation and the specified initial conditions, creating a singular path within the family of general solutions.