Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}-5 x y^{\prime}+9 y=0\)

Short answer

Question: Find the general solution for the given differential equation: \(x^{2} y^{\prime \prime}-5 x y^{\prime}+9 y=0\) Answer: The general solution for the given differential equation is \(y(x) = C_1 x^3 + C_2 x^3 \ln{x}\), valid in any interval not including the singular point (\(x=0\)).

Step-by-step solution

Identify the type of given differential equation

The given differential equation is: \(x^{2} y^{\prime \prime}-5 x y^{\prime}+9 y=0\) This is a second-order linear differential equation with variable coefficients. Notice that the coefficients are powers of \(x\). Thus, we can use the Cauchy-Euler method to find the general solution.

Apply the Cauchy-Euler method

The Cauchy-Euler method replaces the derivatives of the dependent variable \(y\) with the following substitutions: \(y^{\prime \prime} \rightarrow m(m-1)x^{m-2}\) \(y^{\prime} \rightarrow mx^{m-1}\) Substituting these expressions into the given equation, we get: \(x^{2}(m(m-1)x^{m-2}) - 5x(mx^{m-1}) + 9(x^m) = 0\) Simplifying the equation, we get: \(m(m-1)x^m - 5mx^m + 9x^m = 0\)

Find the characteristic equation

Factor out common terms (\(x^m\)) from the simplified equation to obtain the characteristic equation: \(x^m [m(m-1) - 5m + 9] = 0\) Setting the expression inside the brackets equal to zero gives the characteristic equation: \(m(m-1) - 5m + 9 = 0\)

Solve the characteristic equation

Solve the characteristic equation for \(m\): \(m^2 - m - 5m + 9 = m^2 - 6m + 9 = (m - 3)^2 = 0\) So, \(m_1 = 3\) and \(m_2 = 3\) are the roots of the characteristic equation, which are repeated roots.

Write the general solution

For a Cauchy-Euler equation with repeated roots, the general solution will have the form: \(y(x) = C_1 x^{m_1} + C_2 x^{m_2} \ln{x}\) Since \(m_1 = m_2 = 3\), the general solution is: \(y(x) = C_1 x^3 + C_2 x^3 \ln{x}\) This is the general solution of the given differential equation that is valid in any interval not including the singular point (\(x=0\)).

Key concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is an equation involving the unknown function, its derivatives up to the second order, and linear terms of those derivatives. In the context of differential equations, "linear" means that each term either contains the dependent variable or one of its derivatives, raised to the first power (no higher powers or products of these are allowed).

The standard form of a second-order linear differential equation is given by:

• \( a(x) y'' + b(x) y' + c(x) y = g(x) \)

where:

• \(y''\) is the second derivative of \(y\)
• \(y'\) is the first derivative of \(y\)
• \(y\) is the dependent variable
• \(a(x), b(x),\) and \(c(x)\) are functions of the independent variable \(x\)
• \(g(x)\) is the non-homogeneous part

In the case of the equation \(x^{2} y'' - 5x y' + 9y = 0\), \(a(x)\), \(b(x)\), and \(c(x)\) are powers of \(x\), making the use of the Cauchy-Euler method appropriate. Recognizing this structure is vital for applying the correct solution technique.

Variable Coefficients

Variable coefficients in a differential equation are coefficients that are functions of the independent variable rather than constants. When these coefficients depend on \(x\), it makes solving the differential equation more complex compared to constant coefficients.

For our equation, the terms \(x^2, -5x\), and \(9\) are the variable coefficients, indicating that they change as \(x\) changes. This suggests a relationship that isn't constant and might involve transformations depending on \(x\).

The presence of variable coefficients in the Cauchy-Euler form often necessitates the use of specialized methods, such as substitution techniques, to transform the equation into a more manageable form. The recognition of such coefficients helps direct the choice of solution methods, making it possible to find solutions in equations that might otherwise be unapproachable.

Characteristic Equation

The characteristic equation is the algebraic equation that we derive when solving linear differential equations using methods like the Cauchy-Euler approach. This equation is crucial because it helps us determine the roots that lead to the general solution of the differential equation.

For our problem, substituting derivatives based on the format \(y = x^m\), the differential equation simplifies to a polynomial form given by:

• \(m(m-1)x^m - 5mx^m + 9x^m = 0\)

Dropping the \(x^m\) factor, which inherently holds no singular points other than \(x=0\), we solve:

• \(m(m-1) - 5m + 9 = 0\)

This derives the characteristic equation \((m-3)^2 = 0\), which effectively becomes crucial in understanding the behavior of the system described by the differential equation. By solving it, we identify the roots \(m_1 = 3\) and \(m_2 = 3\), which tell us how the solution behaves over different intervals of \(x\).

Repeated Roots Solution

In the context of solving differential equations, repeated roots occur when the characteristic equation yields the same root more than once. In our exercise, the root \(m = 3\) appears twice.

Understanding repeated roots is important because they modify the form of the general solution. For a Cauchy-Euler equation like ours with repeated roots, the general solution generally takes the form:

• \(y(x) = C_1 x^m + C_2 x^m \ln{x}\)

where:

• \(C_1\) and \(C_2\) are constants determined by boundary or initial conditions

Inserting the repeated root value, the solution becomes \(y(x) = C_1 x^3 + C_2 x^3 \ln{x}\). This form is specific to handling the slight nuances of repeated roots, where the logarithmic term is critical in preventing solutions from becoming redundant. Repeated roots solutions extend the spectrum of possible solutions, embracing scenarios where typical methods might fall short.