Problem 23 - Advanced Engineering Mathematics

Question

$$ y^{\prime \prime}+y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = c_1 \cos(t) + c_2 \sin(t) \).

Step 1: Identify the Type of Differential Equation

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

Step 2: Write the Characteristic Equation

For a second-order equation of the form \( ay'' + by' + cy = 0 \), the characteristic equation is \( ar^2 + br + c = 0 \). Since \( a = 1 \), \( b = 0 \), and \( c = 1 \) here, the characteristic equation becomes \( r^2 + 1 = 0 \).

Step 3: Solve the Characteristic Equation

Solve \( r^2 + 1 = 0 \). This equation gives \( r^2 = -1 \). Taking the square root of both sides, we get \( r = i \) or \( r = -i \), where \( i \) is the imaginary unit.

Step 4: Write the General Solution

The roots \( r = i \) and \( r = -i \) are complex and conjugates. For roots of the form \( \, \alpha \pm \beta i \) where \( \, \alpha = 0 \) and \( \, \beta = 1 \), the general solution is \( y(t) = c_1 \cos(t) + c_2 \sin(t) \).

Key Concepts

Second-Order Differential Equation

A second-order differential equation is a type of equation involving a function and its derivatives up to the second derivative. These equations frequently appear in various scientific fields such as physics, engineering, and math. They model systems where the rate of change of a rate of change is significant.

Second-order differential equations typically have the form:

• \( a \, y'' + b \, y' + c \, y = 0 \)

where \( y \) is the function of interest, \( y'' \) is its second derivative, and \( a, b, \) and \( c \) are constants. There might not be a \( y' \) term in every case, as seen in our given equation, \( y'' + y = 0 \).

This equation is considered linear and homogeneous because:

• Linear: It involves no powers or products of \( y \), \( y' \), or \( y'' \).
• Homogeneous: Every term contains a part of the function or its derivatives, and there is no free-standing constant.

Characteristic Equation

Solving a second-order linear differential equation often involves finding the characteristic equation. This equation helps determine the nature of the solution by way of its roots.

For a given second-order differential equation of the form \( a \, y'' + b \, y' + c \, y = 0 \), the corresponding characteristic equation is given by:

• \( ar^2 + br + c = 0 \)

In our specific example, \( a = 1 \), \( b = 0 \), and \( c = 1 \). The characteristic equation simplifies to \( r^2 + 1 = 0 \).

The characteristic equation is a vital step because it transforms the differential equation problem into an algebraic one, making it easier to identify the behavior and type of solutions.

Complex Roots

When solving the characteristic equation, the roots can be real or complex. Complex roots often indicate oscillatory solutions, which can model systems like vibrations or waves.

For the equation \( r^2 + 1 = 0 \), solving gives us complex roots \( r = i \) and \( r = -i \), where \( i \) is the imaginary unit defined by \( i^2 = -1 \). Each root represents a distinct solution component that contributes to the overall behavior of the system.

When the roots are complex, as \( \alpha \pm \beta i \), the general solution to the differential equation becomes:

• \( y(t) = c_1 \cos(\beta t) + c_2 \sin(\beta t) \)

Here, \( \alpha = 0 \) and \( \beta = 1 \), leading to the solution:

• \( y(t) = c_1 \cos(t) + c_2 \sin(t) \)

The constants \( c_1 \) and \( c_2 \) are determined by initial conditions or boundary values, providing flexibility to match specific scenarios or requirements.