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Magazine Chapter 4: Problem 40
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. \(x^{3} \frac{d^{3} y}{d x^{3}}-3 x^{2} \frac{d^{2} y}{d x^{2}}+6 x \frac{d y}{d x}-6 y=3+\ln x^{3}\)
Short Answer
Expert verified
Transforming \( x = e^t \), solve the constant coefficient equation, find general solution for \( y(x) \).
Step by step solution
01
Substitute Variable
We begin by using the substitution \( x = e^t \). Then, \( dx = e^t \, dt \), which implies that \( dt = \frac{dx}{e^t} \) and thus, \( \frac{dx}{dt} = e^t \). Accordingly, we have:\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{e^t} \cdot \frac{dy}{dt} \] This leads to transforming derivative terms in the differential equation.
02
Transform Derivatives
The derivatives transform as follows:1. \( \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dt} \cdot \frac{1}{e^t} \right) \)2. \( \frac{d^3 y}{dx^3} = \frac{d}{dx} \left( \frac{d^2 y}{dx^2} \right) \)After simplifying these derivatives, they become equations with respect to \( t \). Calculating the derivatives:- \( \frac{d^2 y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dt} \cdot \frac{1}{e^t} \right) \cdot \frac{dt}{dx} \)- \( \frac{d^3 y}{dx^3} = \frac{d^2}{dt^2} \left( y \right) \cdot \left( \frac{1}{e^t} \right)^2 \) These simplify using the chain rule.
03
Substitute Back and Simplify
Using the transformed derivatives back into the original equation:\[ (e^t)^3 \left( \frac{d^3 y}{dt^3} \right)e^{-3t} - 3(e^t)^2 \left( \frac{d^2 y}{dt^2} \right)e^{-2t} + 6e^t \left( \frac{dy}{dt} \right)e^{-t} - 6y = 3 + \ln(e^t)^3 \] This simplifies to:\[ \frac{d^3 y}{dt^3} - 3 \frac{d^2 y}{dt^2} + 6 \frac{dy}{dt} - 6y = 3t + 3 \]
04
Solve Constant Coefficient Differential Equation
Now we solve the differential equation:\[ \frac{d^3 y}{dt^3} - 3 \frac{d^2 y}{dt^2} + 6 \frac{dy}{dt} - 6y = 3t + 3 \]Finding the homogenous solution involves solving the characteristic polynomial \( r^3 - 3r^2 + 6r - 6 = 0 \). Solve for \( r \) to find distinct real roots or a repeated root.
05
Find Particular Solution
Using the method of undetermined coefficients, assume a particular solution of the form \( y_p = At + B \) because of the non-homogeneous term \( 3t + 3 \). Substitute into the transformed equation and determine coefficients \( A \) and \( B \) such that the equation is satisfied.
06
Combine Solutions and Transform Back
Combine the homogeneous and particular solutions to obtain the general solution in terms of \( t \). Transform the solution back to \( x \) using \( t = \ln x \). Write the general solution in the original variable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
In solving certain differential equations, especially Cauchy-Euler equations, substituting variables is a powerful technique. Here, we use the substitution \( x = e^t \). This change of variable simplifies the original problem by transforming it into a differential equation with constant coefficients, which is much easier to handle.
- By substituting \( x = e^t \), we transform complex terms involving powers of \( x \) into exponential terms involving \( t \).
- This substitution helps us to systematically transform derivatives with respect to \( x \) into derivatives with respect to \( t \), helping us arrive at an equation manageable with standard methods for constant coefficients.
Differential Equation with Constant Coefficients
Once we perform the substitution \( x = e^t \), the original Cauchy-Euler equation is turned into an equation with constant coefficients. Constant coefficients mean that the coefficients of each derivative term are constants instead of functions of \( x \) or \( t \).
- This transformation is crucial because it allows us to apply various algebraic techniques, such as solving for roots of characteristic polynomials.
- The characteristic equation is derived from the transformed differential equation, and it typically appears as a polynomial equation in terms of \( r \), where \( r \) represents potential solutions for characteristic roots.
Non-Homogeneous Differential Equation
The equation after transformation becomes non-homogeneous due to terms like \( 3t + 3 \), which do not disappear or become zero. A non-homogeneous differential equation contains terms that are added to or subtracted from the zero-value homogeneous counterpart.
- The general solution to such an equation is composed of a complementary (homogeneous) solution and a particular solution.
- The complementary solution is found by solving the equation with zero as its non-homogeneous part.
Method of Undetermined Coefficients
To solve the non-homogeneous differential equation obtained after substitution, we use the method of undetermined coefficients. This technique involves assuming a form for the particular solution based on the non-homogeneous part.
- Typically, you choose polynomial, exponential, or trigonometric functions. Here, because the non-homogeneous term is \( 3t + 3 \), we assume a solution of the form \( y_p = At + B \).
- Substituting this assumed form into the differential equation allows us to solve for \( A \) and \( B \), thus providing a particular solution that satisfies the equation.
