Problem 6 - Advanced Engineering Mathematics

Question

Solve the given differential equation. $$ x^{2} y^{\prime \prime}+5 x y^{\prime}+3 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 x^{-3} + C_2 x^{-1} \).

Step 1: Identify the Differential Equation Type

The given equation is a second order linear homogeneous differential equation with variable coefficients. It is in the form of \( x^{2} y'' + 5x y' + 3y = 0 \).

Step 2: Rewrite in Standard Form

Divide through by \( x^2 \) to make the differential equation in the standard form: \[ y'' + \frac{5}{x} y' + \frac{3}{x^2} y = 0 \].

Step 3: Recognize it as an Euler-Cauchy Equation

The equation \( y'' + \frac{5}{x} y' + \frac{3}{x^2} y = 0 \) is an Euler-Cauchy equation, which can be solved using the substitution \( y = x^m \).

Step 4: Substitute and Simplify

Substitute \( y = x^m \), which gives \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \). Substitute these into the equation:\[ m(m-1)x^{m-2} + \frac{5m}{x} x^{m-1} + \frac{3}{x^2} x^m = 0 \]. Simplifying, the equation becomes:\[ m(m-1) + 5m + 3 = 0 \].

Step 5: Solve the Characteristic Equation

The characteristic equation is \( m(m-1) + 5m + 3 = 0 \). Simplifying gives \( m^2 + 4m + 3 = 0 \). Factor this quadratic equation to get \( (m+3)(m+1) = 0 \). Thus, the solutions are \( m = -3 \) and \( m = -1 \).

Step 6: Write the General Solution

Since we have distinct real roots \( m_1 = -3 \) and \( m_2 = -1 \), the general solution of the Euler-Cauchy equation is:\[ y(x) = C_1 x^{-3} + C_2 x^{-1} \], where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Second Order Differential Equations

A second order differential equation is a type of differential equation in which the highest derivative is the second derivative of the unknown function. In mathematical terms, these equations involve up to the second derivative, denoted as \( y'' \).
This category can often be seen in applications like physics and engineering, where dynamic systems, such as oscillations and wave phenomena, are modeled.
In its general form, a second order differential equation can be represented as:

• \( a(x) y'' + b(x) y' + c(x) y = f(x) \)

Here, \( a(x), b(x), \) and \( c(x) \) are coefficients which may themselves be functions of \( x \). The term \( f(x) \) represents an external function or source term and can affect the equation's "homogeneity" status, as we will explore next.

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms involve the unknown function or its derivatives only. Thus, there is no external term \( f(x) \) added, leading to the equation being set to zero.
For example, the Euler-Cauchy type equation \( x^2 y'' + 5x y' + 3y = 0 \) is homogeneous because there is no term like \( f(x) \).
Homogeneous equations are often easier to solve because they rely on finding solutions as functions of the independent variable (in this case, \( x \)). These solutions typically involve constants, as any linear combination of solutions to a homogeneous equation is also a solution.
The absence of \( f(x) \) implies that these equations describe systems at equilibrium or in natural states without external forces.

Characteristic Equation

The characteristic equation is a critical tool used to solve linear differential equations with constant coefficients, especially in the Euler-Cauchy form.
This equation emerges when we substitute assumed exponential solutions (like \( y = x^m \) for Euler-Cauchy) into the original differential equation.
For instance, substituting \( y = x^m \) into our equation led to the characteristic equation:

• \( m(m-1) + 5m + 3 = 0 \)

Solving the characteristic equation often involves factoring a quadratic expression. In our example, it simplifies and factors as \( (m+3)(m+1) = 0 \), yielding the roots \( m = -3 \), \( m = -1 \).
The roots signify the powers of \( x \) in the general solution, dictating how solutions grow or decay as \( x \) changes.

General Solution

After solving the characteristic equation, the final step is formulating the general solution of the differential equation.
This solution incorporates all potential solutions by integrating the distinct root findings from the characteristic equation. Specifically for distinct real roots \( m_1 \) and \( m_2 \), the general solution of the Euler-Cauchy equation becomes:

• \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \)

Where \( C_1 \) and \( C_2 \) are constants determined by specific initial or boundary conditions, if given.
In our case, the general solution derived from \( m = -3 \) and \( m = -1 \) was \( y(x) = C_1 x^{-3} + C_2 x^{-1} \).
This succinct form encapsulates all possible behaviors of the system as described by the differential equation.