Problem 46 - Advanced Engineering Mathematics

Question

In Problems 43-48, 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 is \( y(x) = C_1 x^2 + C_2 x^3 + \frac{1}{3}\ln(x) + \frac{5}{18} \).

Step 1: Apply the Substitution

To start, apply the substitution \( x = e^t \). This implies that \( dx = e^t \, dt \). Substitute \( x = e^t \) into the equation to transform the Cauchy-Euler equation into a differential equation with constant coefficients.

Step 2: Derivatives Transformation

Using the substitution \( x = e^t \), express \( y' \) and \( y'' \) in terms of derivatives with respect to \( t \).\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} \]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{x} \frac{dy}{dt}) = \frac{d}{dt}(\frac{1}{x} \frac{dy}{dt}) \cdot \frac{dt}{dx} = \frac{1}{x^2} \cdot (\frac{d^2y}{dt^2} - \frac{dy}{dt}) \]Use these transformations in the original equation.

Step 3: Substitute and Simplify

Substitute the transformed derivatives back into the original equation:\[ x^2 \cdot \frac{d^2y}{dt^2} - x \cdot \frac{dy}{dt} - 4x \cdot \frac{dy}{dt} + 6y = 2t \]Simplify the equation to:\[ \frac{d^2y}{dt^2} - 5 \frac{dy}{dt} + 6y = 2t \]This equation now has constant coefficients.

Step 4: Solve the Homogeneous Equation

Find the complementary solution by solving the homogeneous part of \( \frac{d^2y}{dt^2} - 5 \frac{dy}{dt} + 6y = 0 \).Assume a solution \( y = e^{mt} \), then the characteristic equation is:\[ m^2 - 5m + 6 = 0 \]Factor this to find \( m = 2 \) and \( m = 3 \). Therefore, the complementary solution is:\[ y_c = C_1 e^{2t} + C_2 e^{3t} \]

Step 5: Solve the Non-Homogeneous Equation

Use the method of undetermined coefficients to solve:\[ \frac{d^2y}{dt^2} - 5 \frac{dy}{dt} + 6y = 2t \]Assume a particular solution of the form \( y_p = At + B \).Substitute \( y_p \) into the equation to determine \( A \) and \( B \):\[ 0 - 5A + 6At + 6B = 2t \]Match coefficients: \( 6A = 2 \) gives \( A = \frac{1}{3} \) and \( -5A + 6B = 0 \) gives \( B = \frac{5}{18} \). Thus, \( y_p = \frac{1}{3}t + \frac{5}{18} \).

Step 6: General Solution in Terms of t

Combine the complementary and particular solutions:\[ y(t) = C_1 e^{2t} + C_2 e^{3t} + \frac{1}{3}t + \frac{5}{18} \]

Step 7: Translate Solution Back to x

Express the solution in terms of \( x \) by substituting \( x = e^t \), so \( t = ln(x) \):\[ y(x) = C_1 x^2 + C_2 x^3 + \frac{1}{3}\ln(x) + \frac{5}{18} \]

Key Concepts

Constant Coefficients Differential Equation

When working with differential equations, particularly the Cauchy-Euler type, finding an easier way to solve them involves transforming them into what we call differential equations with constant coefficients.
This transformation simplifies the problem significantly. It involves substituting variables and derivatives in such a way that the coefficients of the derivatives become constants rather than functions of the variable itself.
This process results in an equation that is much more straightforward to handle than the original, especially since we have well-established methods and solutions for such equations.

Method of Undetermined Coefficients

In solving non-homogeneous differential equations where the right side isn't zero, the method of undetermined coefficients is quite handy. This technique involves guessing a form for the particular solution based on the form of the non-homogeneous part of the equation.
If the right-hand side of the equation involves functions like constants, polynomials, exponentials or sines and cosines, we can conjecture a similar form for our particular solution.
For example, if it's a polynomial like \(2t\), the guessed solution would be another polynomial.

• Set up the particular solution with unknown coefficients.
• Substitute it back into the equation to solve for these coefficients.

The idea is to balance the equation by determining the values of these coefficients, making this guess an actual solution with components matching the non-homogeneous part.

Characteristic Equation

The characteristic equation is a crucial concept when dealing with second-order linear differential equations with constant coefficients.
This equation provides a way to find the complementary solution of the differential equation, which represents the part of the solution due to the homogeneous nature of the problem.
To form the characteristic equation, assume a solution of the form \(y = e^{mt}\), where \(m\) is a constant to be determined.
Then, substitute this guess into the homogeneous form of the differential equation and simplify to derive a polynomial equation in terms of \(m\).
This polynomial is known as the characteristic equation.
Solving this gives the roots (values of \(m\)) that will help build the complementary solution using exponentials of these roots.

Complementary Solution

The complementary solution refers to the component of the solution to a differential equation that addresses the equation's homogeneous form.
In our context, it deals with solving \(rac{d^2y}{dt^2} - 5rac{dy}{dt} + 6y = 0\).
After finding the roots of the characteristic equation, they inform us how to construct the complementary solution.
If \(m_1\) and \(m_2\) are the roots, whether they are real or complex, we use these to construct terms involving exponentials or sinusoidal functions.

• For distinct real roots \(m_1\) and \(m_2\), the solution is \(y_c = C_1e^{m_1t} + C_2e^{m_2t}\).
• If there are repeated roots, the solution takes a modified form to account for the repetition.

The complementary solution thus forms the base or starting point for finding the general solution of the differential equation when we later add a particular solution to account for the non-homogeneous part.