Problem 36 - A First Course in Differential Equations with Modeling Applications

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 \(4.3-4.5\) $$x^{3} y^{m}-3 x^{2} y^{\prime \prime}+6 x y^{\prime}-6 y=3+\ln x^{3}$$

Step-by-Step Solution

Answer

Substitute \(x = e^t\), then solve the constant coefficient differential equation and back-substitute\.\) Solution: Combine complementary and particular solutions and translate back with \(x = e^t\).

Step 1: Identify the Substitution and Differentiate

We are given the substitution \(x = e^t\). To rewrite the equation in terms of \(t\), we need to express derivatives \(y'\) and \(y''\) in terms of \(t\). The relationship is \(dx = e^t \, dt\), hence \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt} \cdot \frac{1}{e^t}\) and \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dt}\cdot\frac{1}{e^t}\right)\cdot\frac{1}{e^t}\).

Step 2: Rewrite the Derivative Relations

By differentiating we find: \(\frac{dy}{dx} = e^{-t}\frac{dy}{dt}\), \(\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)\). Substitute these into the original equation.

Step 3: Substitute into Original Equation

Substitute the expressions for \(y'\) and \(y''\) into the equation: \(x^3 y^{(m)} - 3x^2 \left(e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)\right) + 6x \left(e^{-t}\frac{dy}{dt}\right) - 6y = 3 + \ln x^3\). Replace \(x = e^t\).

Step 4: Simplify the Equation

Simplify the terms using \(x = e^t\): \(e^{3t} y^{m} - 3 e^{2t} \left(e^{-2t} \frac{d^2y}{dt^2} - e^{-2t} \frac{dy}{dt}\right) + 6 e^t \left(e^{-t}\frac{dy}{dt}\right) - 6y = 3 + 3t\). This simplifies to \(y^{m} - 3\frac{d^2y}{dt^2} + 3\frac{dy}{dt} - 6y = 3 + 3t\). This is a differential equation with constant coefficients.

Step 5: Solve the Transformed Constant Coefficient Equation

Find the complementary solution \(y_c\) by solving \(y^{m} - 3\frac{d^2y}{dt^2} + 3\frac{dy}{dt} - 6y = 0\). The characteristic equation is \(m^3 - 3m^2 + 6m - 6 = 0\). Solve this using techniques from algebra (such as factoring or synthetic division).

Step 6: Solve for Particular Solution

Given the non-homogeneous part \(3 + 3t\), find the particular solution \(y_p\). Assume a solution form based on the non-homogeneous term, such as \(A + Bt\), and substitute it into the differential equation to find constants \(A\) and \(B\).

Step 7: General Solution of Transformed Equation

The general solution of the differential equation is the sum of the complementary and particular solutions: \(y(t) = y_c + y_p\). Integrate and simplify as necessary.

Step 8: Back-Substitute x=e^t

Return to the original variables by substituting back \(x = e^t\), so our solution becomes \(y(x) = y(t=\ln x)\). Replace \(t\) with \(\ln x\) in the general solution.

Key Concepts

Cauchy-Euler Equation

The Cauchy-Euler equation is a specific type of linear differential equation. It often appears in the form of

• \( a_0 x^n y^{(n)} + a_1 x^{n-1} y^{(n-1)} + \ldots + a_n y = f(x) \),

where the coefficients are powers of \( x \). This equation is particularly useful in engineering and physics where problems can have polynomial behavior.
A unique characteristic of the Cauchy-Euler equation is its dependency on the powers of the independent variable \( x \). This property can make solving it more challenging than equations with constant coefficients.
However, a clever change of variables or substitution can transform it into a more manageable form, making the solution process much simpler.

Substitution Method

The substitution method is a powerful technique for solving differential equations, especially useful when dealing with Cauchy-Euler equations. In our exercise, we utilize the substitution \( x = e^t \), allowing us to transform the equation into a function of \( t \), where differentiation and integration can be more straightforward.
This transformation leverages the relationship between \( x \) and \( t \), where \( t = \ln x \) and \( dx = e^t \, dt \). This change simplifies the equation by eliminating the variable \( x \), leading to differential equations with constant coefficients.

• Express \( y' \) and \( y'' \) in terms of \( t \),
• Substitute back to solve the simplified equation.

By reducing the equation complexity, substitution provides an efficient pathway to finding solutions.

Constant Coefficients

Differential equations with constant coefficients are simpler forms. Once the original equation is transformed using a substitution method, it often yields an equation where these coefficients remain constant. This is where constant coefficients come into play.
In this context, the differential equation takes a more standard form, allowing us to use classical solution techniques:

• Find the characteristic equation, derived from setting the differential equation to zero.
• Solve this equation, typically a polynomial, to find solutions for \( m \).

These solutions form the basis of the complementary solution, which captures the homogeneous part of our transformed equation.

Particular Solution

A particular solution adds to the complementary solution to account for any non-homogeneous components of the differential equation. It specifically addresses the extra terms that are not captured by the homogeneous equation alone.
For the non-homogeneous term in our exercise, \( 3 + 3t \), we propose forms like \( A + Bt \) that mirror these terms. Then, by plugging this assumption into the equation, we determine the constants \( A \) and \( B \) that satisfy the equation under consideration.
This approach ensures that all aspects of the equation, both homogeneous and non-homogeneous, are appropriately addressed. Once both parts are solved, they are combined to formulate the general solution.