Problem 47 - 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. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+13 y=4+3 x $$

Step-by-Step Solution

Answer

The solution is \( y(x) = x^2 \left(C_1\cos(3\ln x) + C_2\sin(3\ln x)\right) + \frac{4}{13} + \frac{3}{5}x \).

Step 1: Substitute Variable

We begin by substituting \(x = e^t\), which implies that \( rac{dx}{dt} = e^t \) and \( rac{dt}{dx} = \frac{1}{e^t} = \frac{1}{x}\). This means \( y'(x) = \frac{dy}{dx} = \frac{dy}{dt}\cdot \frac{dt}{dx} = \frac{1}{x}\cdot\frac{dy}{dt} \). Similarly, \( y''(x) = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1}{x}\cdot\frac{dy}{dt}\right) = \frac{1}{x^2}\frac{d^2y}{dt^2} - \frac{1}{x^2}\cdot\frac{dy}{dt}\).

Step 2: Transform the Equation

With these derivatives, the original equation \( x^2y'' - 3xy' + 13y = 4 + 3x \) is transformed. Substituting into this equation gives us \( y'' - \frac{1}{x}y' - 3y' + 13y = 4 + 3e^t \) because \( x \cdot \frac{1}{x} \cdot \left(y'\right) \) simplifies to \(y'\). Simplify the transformed expression.

Step 3: Simplify and Rearrange

Rewriting the equation, we have \( y'' - 4y' + 13y = 4 + 3e^t \). This is now a differential equation with constant coefficients.

Step 4: Solve the Homogeneous Equation

The associated homogeneous equation is \( y'' - 4y' + 13y = 0 \). The characteristic equation is \( r^2 - 4r + 13 = 0 \). Solving this quadratic gives us \( r = 2 \pm 3i \). This means the general solution to the homogeneous equation is \( y_h(t) = e^{2t}(C_1\cos(3t) + C_2\sin(3t)) \).

Step 5: Solve the Non-homogeneous Equation

The particular solution involves solving \( y'' - 4y' + 13y = 4 + 3e^t \). Assuming a particular solution of form \( y_p(t) = A + Be^t \), compute derivatives, substitute into equation, and solve for \( A \) and \( B \). First part yields constant \( A = \frac{4}{13} \). Next, substituting and equating coefficients yields \( B = \frac{3}{5} \). Thus, \( y_p(t) = \frac{4}{13} + \frac{3}{5}e^t \).

Step 6: Form the General Solution

The general solution \( y(t) \) is the sum of the homogeneous solution and the particular solution: \[y(t) = e^{2t}(C_1\cos(3t) + C_2\sin(3t)) + \frac{4}{13} + \frac{3}{5}e^t\].

Step 7: Transform Back to Original Variables

Substitute back \( t = \ln x \) (because \( x = e^t \)) into the general solution, giving:\[y(x) = x^2 \left(C_1\cos(3\ln x) + C_2\sin(3\ln x)\right) + \frac{4}{13} + \frac{3}{5}x\].

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They play a crucial role in modeling various natural processes and phenomena such as motion, growth, and decay.
In the exercise, we focus on a specific type known as the Cauchy-Euler differential equation. This equation arises in problems involving exponential functions and power series. The key feature of a Cauchy-Euler equation is the variable coefficients that depend on powers of the independent variable.
To simplify solving, we often transform these equations into those with constant coefficients, allowing us to apply standard solution methods effectively. Understanding differential equations is vital for fields like physics, engineering, and economics, as they provide the tools for predicting and understanding behavior in dynamic systems.

Constant Coefficients

Constant coefficients in differential equations mean that the coefficients of the terms in the equation do not change. They are fixed values, usually numbers or known constants.
Having constant coefficients allows for solving differential equations more easily using algebraic methods. One common technique is transforming a Cauchy-Euler equation into one with constant coefficients through substitution methods.
In this exercise, we substitute variables to change our original equation into a much more manageable form. This transformed equation then opens the door to straightforward algebraic techniques, such as finding characteristic equations, to solve it.
Constant coefficients simplify complex problems, making them an integral part of studying differential equations.

Particular Solution

A particular solution in the context of differential equations is a specific solution that satisfies not just the homogeneous part of the equation but also the entire non-homogeneous equation. The non-homogeneous equation includes an additional function, like detailed in our exercise, such as "4 + 3x."
The process of finding a particular solution typically involves making educated guesses about the form of the solution based on the type of the non-homogeneous part, which could be constants, exponentials, sines, cosines, etc.
In our solution, we assume a particular solution of the form \(y_p(t) = A + Be^t\), calculate its derivatives, and substitute it into the transformed equation to isolate and determine the values of \(A\) and \(B\).
These calculated values plug back into our assumed form to give the particular solution, addressing the unique aspects of our differential equation beyond the general homogeneous solution.

Characteristic Equation

The characteristic equation is a crucial step in solving differential equations, specifically those with constant coefficients. It arises from the homogeneous part of a differential equation by assuming the solution is of an exponential form.
For example, with our transformed equation \(y'' - 4y' + 13y = 0\), we assume solutions of the form \(e^{rt}\). Substituting these into the equation and setting up the result as a polynomial gives us the characteristic equation.
Solving this polynomial reveals the roots, which tell us about the behavior of the solutions. Real roots lead to exponential solutions, while complex roots, like in our exercise where \(r = 2 \pm 3i\), give rise to oscillatory solutions in the form of sines and cosines mixed with exponentials.
The characteristic equation is a pillar in the theory of linear differential equations, granting insights into solution forms and allowing us to construct the general solutions effectively.