91Ó°ÊÓ

A homogeneous solution refers to the solution of a differential equation when the equation equals zero. In our exercise, the equation looks like this:

• \( \frac{d^2 y}{dx^2} + y = 0 \).

This form is due to ignoring the non-homogeneous part (\(\sec^2 x\)). Solving the homogeneous equation involves finding solutions that match just the structure of the differential operator.To solve, we use the characteristic equation. This is done by substituting a trial solution like \(y = e^{rx}\) into the equation, leading to what's called the characteristic equation - \(r^2 + 1 = 0\). The solutions for this are \(r = i\) and \(r = -i\), indicating complex roots.With complex roots, the homogeneous solution takes the form:

• \(y_c = C_1 \, \cos x + C_2 \, \sin x \).

Here, \(C_1\) and \(C_2\) are arbitrary constants to be determined later, usually by initial conditions or constraints.

The particular solution is a solution to the full non-homogeneous differential equation. In our case, for \(\frac{d^2 y}{dx^2} + y = \sec^2 x\), we need to account for the non-zero "right-hand side" of the equation.A common way to discover a particular solution is through inspection or using known methods like undetermined coefficients or variation of parameters. In this problem, the proposed particular solution was \(y_p = \tan x\). The form is guessed based on our understanding of the \(\sec^2 x\) term, since \(\frac{d}{dx}(\tan x) = \sec^2 x\).By substituting \(y_p = \tan x\) back into the original equation, we can verify if it satisfies the non-homogeneous equation.

• If the substitution works (as it does in this case), then \(y_p = \tan x\) is indeed a valid particular solution.

The full solution to the differential equation is a combination of the homogeneous and particular solutions.

Initial conditions are essential for finding a specific solution to the differential equation tailored to a scenario. They are given values for the solution and its derivatives at a particular point.For our exercise, the initial conditions are:

• \( y(0) = 1 \)
• \( y'(0) = 1 \)

These conditions help us solve the constants \(C_1\) and \(C_2\) in the homogeneous solution.Applying these, we substitute back into the full solution:

• \( y(x) = C_1 \cdot \cos x + C_2 \cdot \sin x + \tan x \)
• At \( x = 0 \), \( y(0) = 1 = C_1 \cdot 1 + C_2 \cdot 0 + \tan 0 \), resulting in \(C_1 = 1\).
• The derivative, \( y'(x) = -C_1 \sin x + C_2 \cos x + \sec^2 x \).
• \( y'(0) = 1 = -1 \cdot 0 + C_2 \cdot 1 + 1\), leading to \(C_2 = 0\).

Thus, the specific solution considers all conditions, not just the equation.

The characteristic equation is a crucial tool for solving linear homogeneous differential equations. It lets us find the form of the homogeneous solution.For our equation \(\frac{d^2 y}{dx^2} + y = 0\), the characteristic equation arises from substituting \(y = e^{rx}\) and its derivatives into the homogeneous equation, resulting in an algebraic equation.This leads to:

• \( r^2 + 1 = 0 \)
• The roots here are \( r = i \) and \( r = -i \), which are imaginary.

Imaginary roots tell us the homogeneous solution has oscillatory components:

• \( y_c = C_1 \cos(x) + C_2 \sin(x) \)

This form leverages Euler's formula, showing how complex solutions translate into sines and cosines for real-valued functions.