Problem 11 - Advanced Engineering Mathematics

Question

In Problems 1-18, solve the given differential equation. $$ x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0 $$

Step-by-Step Solution

Answer

The solution is \(y(x) = C_{1} x^{-2} + C_{2} x^{-2} \ln{x}\).

Step 1: Identify the Type of Equation

The differential equation given is \(x^{2} y'' + 5x y' + 4y = 0\). It is a second-order linear differential equation with variable coefficients. Such equations are often solved using special methods such as the method of Frobenius or recognizing them as Euler (Cauchy-Euler) equations.

Step 2: Confirm it as a Cauchy-Euler Equation

Recognize that a Cauchy-Euler equation has the form \(a x^{2} y'' + b x y' + c y = 0\). Our equation matches this form with \(a=1\), \(b=5\), and \(c=4\).

Step 3: Find the Characteristic Equation

For a Cauchy-Euler equation \(a x^{2} y'' + b x y' + c y = 0\), the characteristic equation is \(a r(r-1) + b r + c = 0\). Substitute \(a=1\), \(b=5\), \(c=4\) to find \(r(r-1) + 5r + 4 = 0\).

Step 4: Solve the Characteristic Equation

Simplify the characteristic equation: \(r^{2} + 4r + 4 = 0\). This factors as \((r+2)^{2} = 0\). Hence, \(r = -2\) is a repeated root.

Step 5: Write the General Solution

For repeated roots \(r\), the solution is given by \(y(x) = C_{1} x^{r} + C_{2} x^{r} \ln{x}\). Applying \(r = -2\), the general solution is \(y(x) = C_{1} x^{-2} + C_{2} x^{-2} \ln{x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Second-Order Linear Differential Equation

In the exercise, you are dealing with a second-order linear differential equation. But what does that mean? Let's break it down to its components.

Firstly, a differential equation is an equation that relates a function to its derivatives. It describes how a function changes based on its current state. Now, a linear differential equation is one in which the dependent variable (usually denoted as \( y \)) and its derivatives appear linearly. This means they are not raised to any power other than one, and they are not multiplied together.

The phrase "second-order" refers to the highest derivative in the equation, which in this case is the second derivative \( y'' \). Second-order differential equations often appear in physics and engineering, modeling oscillations, waves, and other phenomena.

When working with a second-order differential equation with variable coefficients, such as our example \( x^{2} y'' + 5x y' + 4y = 0 \), traditional solution techniques used for constant coefficients do not apply. Instead, we must use methods specific to variable coefficients, like the Cauchy-Euler approach.

Characteristic Equation

The characteristic equation is a crucial step in solving Cauchy-Euler differential equations. It's like a special formula that helps us figure out the solution to the differential equation.

For a general Cauchy-Euler equation \( a x^{2} y'' + b x y' + c y = 0 \), the characteristic equation is found by assuming a solution of the form \( y = x^{r} \), where \( r \) is a constant. Substituting this into the differential equation leads to a polynomial equation in terms of \( r \), given by \( a r(r-1) + b r + c = 0 \).

• This equation is derived because it simplifies the process of finding \( y \) by reducing it to solving for \( r \).
• Once \( r \) is found, it gives us insights into the form of the general solution.
• If the characteristic equation has real and distinct roots, the solution is direct. If there's a repeated root, as in the example, the solution takes a special form.

In our problem, the characteristic equation is \( r(r-1) + 5r + 4 = 0 \), which simplifies to \( r^{2} + 4r + 4 = 0 \). Solving, we find \( r = -2 \) with multiplicity, indicating a repeated root. This influences the form of our general solution significantly.

Method of Frobenius

The Method of Frobenius is a powerful technique used to solve differential equations, especially when variable coefficients prevent the use of simpler, more straightforward methods.

It is particularly useful for equations that do not lend themselves easily to solutions through standard power series. Instead, the method assumes a solution in the form of a series expansion: \[ y(x) = x^{r} \sum_{n=0}^{abla} a_n x^{n} \] This form allows for flexibility when dealing with points where the equation might not behave nicely, like singular points.

The Frobenius method is often engaged when working through Cauchy-Euler equations with complex or repeated roots, providing a systematic approach to find solutions.

• It involves substituting the series into the differential equation and matching coefficients to determine the values of \( a_n \) and \( r \).
• Through this method, one can find both the solution at ordinary points and the development of the solution near singular points.

While in our specific exercise the Frobenius method may not be directly applied, it is a valuable technique in the toolbox for more complex situations. It represents the level of depth available when tackling second-order equations with intricate characteristics.