Problem 45 - 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. $$ x^{2} y^{\prime \prime}+10 x y^{\prime}+8 y=x^{2} $$

Step-by-Step Solution

Answer

The solution is \(y(x) = C_1 x^{-1} + C_2 x^{-8} + \frac{1}{30} x^2\).

Step 1: Recognize the Substitution

We're given the Cauchy-Euler equation \(x^2 y'' + 10x y' + 8y = x^2\). We need to use the substitution \(x = e^t\), which implies \(dx/dt = e^t\) or \(dt/dx = 1/x\). The relationship between derivatives is useful to transform the equation.

Step 2: Transform Derivatives

Using the chain rule, express the derivatives with respect to \(t\): \(y' = \frac{dy}{dx} = \frac{1}{x} \frac{dy}{dt} = \frac{1}{e^t} \frac{dy}{dt}\) and \(y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right) = \frac{1}{x^2}\frac{d^2y}{dt^2} - \frac{1}{x^2}\frac{dy}{dt}\). Substitute \(x = e^t\) into these expressions.

Step 3: Substitute and Simplify the Equation

Substitute \(x = e^t\), \(y' = e^{-t} y_t\), and \(y'' = e^{-2t} y_{tt} - e^{-2t} y_t\) into the original equation to transform it: \(e^{2t} (e^{-2t}y_{tt} - e^{-2t} y_t) + 10 e^t (e^{-t} y_t) + 8y = e^{2t}\). Simplify to \(y_{tt} + 9y_t + 8y = e^{2t}\).

Step 4: Solve the Homogeneous Equation

Solve the homogeneous part \(y_{tt} + 9y_t + 8y = 0\). The characteristic equation is \(r^2 + 9r + 8 = 0\), which factors to \((r+1)(r+8) = 0\). Thus, the roots are \(-1\) and \(-8\). The general solution for the homogeneous part is \(y_h = C_1 e^{-t} + C_2 e^{-8t}\).

Step 5: Solve for Particular Solution

To find a particular solution \(y_p\), use the method of undetermined coefficients. Assume \(y_p = Ae^{2t}\). Then \(y_{pt} = 2Ae^{2t}\) and \( y_{ptt} = 4Ae^{2t}\). Substitute these into \(y_{tt} + 9y_t + 8y = e^{2t}\) and solve for \(A\): \(4A + 18A + 8A = 1\) gives \(30A = 1\) so \(A = 1/30\). Thus, \(y_p = (1/30) e^{2t}\).

Step 6: Combine Homogeneous and Particular Solutions

Combine the solutions to form the general solution for \(y(t)\): \(y(t) = y_h + y_p = C_1 e^{-t} + C_2 e^{-8t} + \frac{1}{30} e^{2t}\).

Step 7: Translate Back to Original Variable

Substitute back \(t = \ln(x)\), which gives \(e^t = x\), into the general solution: \(y(x) = C_1 x^{-1} + C_2 x^{-8} + \frac{1}{30} x^2\). This is the solution to the original Cauchy-Euler equation.

Key Concepts

Substitution Method

The substitution method is a powerful technique used mainly to simplify solving differential equations, especially when dealing with Cauchy-Euler equations. The key idea is to substitute a variable that transforms the problem into a more manageable form. In this case, we substitute the variable \( x = e^t \). This particular choice is beneficial as it helps convert a Cauchy-Euler equation into a differential equation with constant coefficients, which are easier to solve.Here's how it works: by substituting \( x = e^t \), we can turn the problem’s non-constant coefficients into constants. We then utilize the chain rule to rewrite derivative expressions in terms of \( t \) rather than \( x \). For example:

• \( y' = \frac{1}{e^t} \frac{dy}{dt} \)
• \( y'' = \frac{d}{dx}(\frac{1}{x} \frac{dy}{dt}) \)

This transformation makes the differential equation more straightforward to solve.

Differential Equations with Constant Coefficients

Once transformed using the substitution \( x = e^t \), a Cauchy-Euler equation turns into a differential equation with constant coefficients. These types of differential equations are much simpler to handle.The general form of a linear differential equation with constant coefficients is:\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = g(t) \]where each \( a_i \) is a constant, and \( g(t) \) is a known function.Solving these equations involves finding both a homogeneous solution, where \( g(t) = 0 \), and a particular solution, where \( g(t) \) is not zero. The sum of these solutions gives the general solution for the differential equation.

Characteristic Equation

The characteristic equation is a polynomial equation derived from the differential equation when we assume solutions of a specific exponential form. For a linear differential equation with constant coefficients, we assume solutions of the form \( y(t) = e^{rt} \).Substituting this assumed solution into the homogeneous differential equation and factoring out \( e^{rt} \) gives us the characteristic equation. For the given problem, transforming the Cauchy-Euler equation resulted in:\[ y_{tt} + 9y_t + 8y = 0 \]which translates to the characteristic equation:\[ r^2 + 9r + 8 = 0 \]Solving this quadratic equation, typically through factoring or the quadratic formula, provides roots that help construct the solution to the homogeneous part of the differential equation.

Undetermined Coefficients

To solve the inhomogeneous differential equation, we utilize the method of undetermined coefficients. This involves guessing a form for the particular solution \( y_p(t) \) based on the right-hand side non-homogeneous term. Here, since the non-homogeneous part is \( e^{2t} \), we propose a solution like \( y_p = A e^{2t} \).We then calculate the derivatives of this guess:

• \( y_{pt} = 2A e^{2t} \)
• \( y_{ptt} = 4A e^{2t} \)

Substituting these into the differentiated equation allows us to solve for the unknown coefficient \( A \). So, for the example, we find:\[ 4A + 18A + 8A = 1 \]Solving this gives \( A = \frac{1}{30} \), leading to the particular solution:\[ y_p = \frac{1}{30} e^{2t} \]This determined coefficient completes our general solution, entwining both the homogeneous and particular solutions to form the overall response to the original equation.