Question

Find the general solution of the given Euler equation on \((0, \infty)\). $$ x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0 $$

Short answer

Question: Find the general solution of the given Euler equation: $$ x^{2}y^{\prime \prime}+5xy^{\prime}+4y=0 $$ Answer: The general solution of the given Euler equation on \((0, \infty)\) is: $$y(x) = C_1x^{-2} + C_2x^{-2}\ln{x}$$ where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Identify the given equation type and apply the substitution

The problem provides us with the Euler equation: $$ x^{2}y^{\prime \prime}+5xy^{\prime}+4y=0 $$ Let's apply the substitution \(y(x)=x^r\), so we'll find the derivatives and replace them in the equation.

Calculate the first and second derivatives

Find the first derivative with respect to \(x\): $$ y'(x) = r x^{r-1} $$ Then, find the second derivative with respect to \(x\): $$ y''(x) = r(r-1)x^{r-2} $$

Replace the derivatives into the given Euler equation

Replace \(y, y', y''\) in the given equation $$ x^2 r(r-1)x^{r-2} + 5x rx^{r-1} + 4x^r = 0 $$

Simplify the Euler equation

Factor the equation to get the characteristic equation $$ x^{r-2}[(r^2 +4r)x^2+4x^r]=0 $$ Since \(x^{r-2}\) is always positive, the factor inside the brackets should be equal to zero: $$ r^2 +4r+ 4=0 $$

Solve the characteristic equation

In this step, we will solve the quadratic equation \( r^2 + 4r + 4 = 0\) to find the values of \(r\). This can be done by factoring or using the quadratic formula: $$ (r+2)^2=0 $$ So, we get only one value for \(r\): $$ r=-2 $$

Construct the general solution

Since there is only one repeated root, r = -2, the solution will take the following form: $$ y(x) = C_1x^{-2} + C_2x^{-2}\ln{x} $$ This is the general form of the solution for the given Euler equation on \((0, \infty)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the realm of calculus, they serve as a tool to describe phenomena where change occurs, such as physics, engineering, and economics. When it comes to specific types, one important category is the Euler or Cauchy-Euler equation. It appears in the forms of a second-order linear differential equation and is distinctive because the coefficients of the terms are powers of the variable, typically denoted as \( x \). In our given exercise, we have such an Euler equation:
\[ x^{2} y^{\textquotesingle \textquotesingle}+5 x y^{\textquotesingle}+4 y=0 \]
This type of differential equation is especially fascinating because it often emerges in problems involving spatial geometries, such as those found in classical mechanics or electromagnetic theory.

Characteristic Equation

A key step in solving linear differential equations is finding the characteristic equation, which is a polynomial equation whose roots help in constructing the general solution to the differential equation. Specifically, for Euler equations, the substitution method can transform the original differential equation into a characteristic equation, which is easier to solve. In the provided exercise, the characteristic equation ascertained after substitution and simplification is:
\[ r^{2} + 4r + 4 = 0 \]
Solving this gives us the roots that are critical to determining the form of the general solution. For this exercise, we only get one root, \( r = -2 \), which appears twice due to the nature of the quadratic.

Methods of Solving Differential Equations

Differential equations can be tackled using various methods, with some being more suitable than others depending on the type of equation one is dealing with. The method of separation of variables, integrating factors, series solutions, and transformation into a characteristic equation are all potent tools in the mathematician's toolkit. In the context of the Euler equation from our exercise, the characteristic equation method applied through a substitution strategy is particularly effective. By substituting \( y(x) = x^{r} \), where \( r \) is an unknown constant, we are able to turn the variable coefficients into a polynomial, which paves the way for us to use algebraic methods to solve for \( r \). The real strength of this method is in its ability to handle the variable coefficients that are typical of Euler equations.

Euler Equation Substitution Method

The Euler equation substitution method involves transforming the original differential equation into a simpler form by making a clever choice of a new unknown function. This substitution hinges on the assumption that solutions to the Euler equation can be represented as \( y(x) = x^{r} \), where the exponent \( r \) is what we're solving for. By calculating the first and second derivatives of \( y \) with respect to \( x \) and substituting them back into the Euler equation, we generate a characteristic equation. This characteristic equation is a standard quadratic equation which can be solved using factoring or the quadratic formula. When we get a repeated root as in our exercise, the general solution includes both the solution for \( r \) and an additional solution that accounts for the multiplicity of the root, involving a natural logarithm term. Therefore, the final form of the solution for our example is:
\[ y(x) = C_{1}x^{-2} + C_{2}x^{-2}\text{ln}{(x)} \]
This general solution constitutes a comprehensive description of the behavior of the differential equation's solutions across its domain.