Problem 46 - Advanced Engineering Mathematics

Question

Use the substitution \(x=e^{t}\) to transform the given Cauchy-Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation using the procedures in Sections 3.3-3.5. $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=\ln x^{2} $$

Step-by-Step Solution

Answer

The solution to the original equation is \( y(x) = C_1 x^2 + C_2 x^3 + \frac{1}{3} \ln x + \frac{5}{18} \).

Step 1: Substitute and Transform the Equation

We begin by substituting \( x = e^t \), which implies \( rac{dx}{dt} = e^t \) and thus \( dt = \frac{1}{x}dx \). When substituted, \( y' = \frac{dy}{dx} \) becomes \( \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} \), and \( y'' = \frac{d}{dx}(y') = \frac{d}{dt}(\frac{dy}{dt} \cdot \frac{1}{x} )\cdot \frac{dt}{dx} = \frac{d^2y}{dt^2} \cdot \frac{1}{x^2} - \frac{1}{x^3} \cdot \frac{dy}{dt} \). This transforms the original equation with substitution.

Step 2: Simplify the Transformed Equation

Substituting these into our differential equation: \( x^{2} y^{\prime\prime} - 4 x y^{\prime} + 6y = \ln(x^2) \), we get \[ y^{\prime\prime}(t) - 5y^{\prime}(t) + 6y(t) = 2t. \] This is a linear differential equation with constant coefficients.

Step 3: Solve the Homogeneous Part

The homogeneous part of the equation is \( y^{\prime\prime} - 5y^{\prime} + 6y = 0 \). Assume a solution of the form \( y(t) = e^{rt} \). Therefore, the characteristic equation is \( r^2 - 5r + 6 = 0 \). Factoring gives \( (r - 2)(r - 3) = 0 \), thus \( r = 2 \) and \( r = 3 \). Hence, the general solution to the homogeneous equation is \( y_h(t) = C_1 e^{2t} + C_2 e^{3t} \).

Step 4: Solve the Particular Part

To solve the particular part \( y^{\prime\prime} - 5y^{\prime} + 6y = 2t \), assume a particular solution of the form \( y_p(t) = At + B \). Calculating derivatives, \( y_p'(t) = A \) and \( y_p''(t) = 0 \). Substituting these into the equation gives \(-5A + 6(At + B) = 2t\). Simplifying results in \( (6A)t + (6B - 5A) = 2t \). Thus, equating coefficients gives \( 6A = 2 \) leading to \( A = \frac{1}{3} \), and \( 6B - 5A = 0 \) resulting in \( B = \frac{5}{18}\). Therefore, \( y_p(t) = \frac{1}{3}t + \frac{5}{18} \).

Step 5: Combine Solutions

The general solution to the transformed equation is \[ y(t) = y_h(t) + y_p(t) = C_1 e^{2t} + C_2 e^{3t} + \frac{1}{3} t + \frac{5}{18}. \]

Step 6: Transform Back to x-domain

Revert the solution back to terms of \( x \) by substituting \( t = \ln x \), which gives \( y(x) = C_1 x^2 + C_2 x^3 + \frac{1}{3} \ln x + \frac{5}{18} \).

Key Concepts

Differential Equation with Constant Coefficients

Differential equations with constant coefficients emerge from scenarios where the coefficients of the derivatives in an equation are constants. This contrasts with Cauchy-Euler equations where coefficients depend on the powers of the variable.
In this exercise, the original Cauchy-Euler form had variable coefficients that depended on powers of \( x \). Through substitution \( x = e^t \), we transitioned these into a form where coefficients became constants, specifically the equation \( y''(t) - 5y'(t) + 6y(t) = 2t \).

This transformation is beneficial as solving differential equations with constant coefficients is more straightforward. Such equations often permit solutions using characteristic equations, which simplifies finding the homogeneous solution.

Transformation Using Substitution

A crucial part of solving Cauchy-Euler equations is the transformation by substitution. We used the technique of substituting \( x = e^t \). This transformation turns a complicated equation with variable coefficients into a more manageable form with constant coefficients. Here's how it works:

• Replace \( x \) with \( e^t \).
• Adjust derivatives: For example, \( y' \) and \( y'' \) were expressed in terms of \( t \) and \( \frac{dy}{dt} \).

This allowed us to systematically break down and transform the original equation into something easier to work with, making the analysis and identification of solutions much easier.

Homogeneous and Particular Solutions

When solving linear differential equations with constant coefficients, we separate the work into finding homogeneous and particular solutions.

1. **Homogeneous Solution**: We solve \( y'' - 5y' + 6y = 0 \) by assuming a solution of the form \( y(t) = e^{rt} \). The characteristic equation helps determine \( r \), leading us to find that \( y_h(t) = C_1 e^{2t} + C_2 e^{3t} \).
2. **Particular Solution**: For the equation \( y'' - 5y' + 6y = 2t \), we assume \( y_p(t) = At + B \). By substituting into the differential equation and matching coefficients, we find \( A \) and \( B \).
Hence, our particular solution is \( y_p(t) = \frac{1}{3}t + \frac{5}{18} \).

Combining these solutions gives the general solution: \( y(t) = y_h(t) + y_p(t) \).

Characteristic Equation in Differential Equations

The characteristic equation is a powerful tool in solving differential equations with constant coefficients. It relates directly to the homogeneous part of the equation, helping to identify solutions quickly.

To find it, assume solutions of the form \( y(t) = e^{rt} \). Substituting into the homogeneous equation \( y'' - 5y' + 6y = 0 \) results in the characteristic equation \( r^2 - 5r + 6 = 0 \).

• Factor the characteristic equation to find \( r \).
• The roots, \( r = 2 \) and \( r = 3 \), help form the solution as \( y_h(t) = C_1 e^{2t} + C_2 e^{3t} \).

This technique turns complex differential problems into algebraic ones, making the problem easier to solve by revealing the order of exponential terms in the solution.