Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}-4 x y^{\prime}+4 y=0\)

Short answer

Question: Find the general solution for the second-order linear homogeneous differential equation \(x^{2} y^{\prime \prime}-4 x y^{\prime}+4 y=0\), valid in any interval not including the singular point. Answer: The general solution for the given differential equation is \(y(x) = \int x^4(4x^{-5}\int x^{-6}y(x) dx + C_1) dx + C_2\), where \(C_1\) and \(C_2\) are constants of integration and the exact form of the integral \(\int x^{-6}y(x) dx\) would depend on more information about the function \(y(x)\).

Step-by-step solution

Identify the type of differential equation

The given equation is a second-order linear homogeneous differential equation with variable coefficients: \(x^{2} y^{\prime \prime}-4 x y^{\prime}+4 y=0\).

Identifying Singular Point(s) of the Differential Equation

In this exercise, we are asked to find the general solution that is valid in any interval not including the singular point. A singular point occurs when the leading coefficient of the highest-order derivative becomes zero. In this case, the leading coefficient is \(x^2\) in front of \(y^{\prime \prime}\). So, we have: A singular point exists when \(x^2 = 0 \implies x = 0\). Therefore, the singular point is x = 0.

Rewriting the differential equation using new variable

By substituting \(p(x)= y'(x)\), our original equation becomes \(x^2 p'(x) - 4xp(x) + 4y(x) = 0\), which is a first-order linear, inhomogeneous differential equation.

Solving the first-order linear, inhomogeneous differential equation for p(x)

Dividing by \(x^2\) yields: \(p'(x) - \frac{4}{x}p(x) + \frac{4}{x^2}y(x) = 0\) To solve this equation, we need to find an integrating factor \(\mu(x)\). The integrating factor \(\mu(x)\) is given by the exponential of the integral of the coefficient of \(p(x)\): \(\mu(x) = e^{\int-\frac{4}{x} \,dx} = e^{-4\ln{x}} = x^{-4}\) Multiplying the equation by the integrating factor \(x^{-4}\), we have: \(x^{-4}p'(x) - 4x^{-5}p(x) + 4x^{-6}y(x) = 0\) This resulting equation is an exact differential equation: \((x^{-4}p(x))' = 4x^{-6}y(x)\) Integrate both sides with respect to x: \(\int(x^{-4}p(x))' dx = \int 4x^{-6}y(x) dx\) Thus, we get: \(x^{-4}p(x) = C_1 - 4x^{-5}\int x^{-6}y(x) dx\) Solve for \(p(x)\): \(p(x) = x^4(4x^{-5}\int x^{-6}y(x) dx + C_1)\)

Obtaining the general solution

Since we made the substitution \(p(x) = y'(x)\), it is time to substitute back and get our general solution for \(y(x)\): \(y'(x) = x^4(4x^{-5}\int x^{-6}y(x) dx + C_1)\) Integrating both sides gives: \(y(x) = \int x^4(4x^{-5}\int x^{-6}y(x) dx + C_1) dx + C_2\) This is the most general form of the solution to the given differential equation, valid in any interval not including the singular point x = 0. Here, \(C_1\) and \(C_2\) are the constants of integration. In general, determining the exact form of the integral \(\int x^{-6}y(x) dx\) would require more information about the function \(y(x)\).

Key concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is a type of differential equation that involves the second derivative of a function. These equations often have the form: \[ a(x) y'' + b(x) y' + c(x) y = 0 \] where \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), and \(y\) is the unknown function we want to solve for. In simpler terms, this equation describes how the rate of change of the rate of change (second derivative) of a quantity relates to the quantity itself and its first derivative.

• "Second-order" refers to the highest derivative, which is the second derivative (\(y''\)).
• "Linear" indicates that \(y\) and its derivatives are to the power of 1.
• "Homogeneous" means the equation equals zero.

Understanding this type of equation is crucial because it frequently appears in physics and engineering, especially when describing systems like oscillations, circuit analysis, and heat conduction.

Singular Points

In the context of differential equations, singular points are values of \(x\) where the equation may become undefined or where the solution behaves differently than expected. They are found by setting the leading coefficient of the highest-order derivative to zero. For the equation \(x^{2}y'' - 4xy' + 4y = 0\), the leading coefficient is \(x^2\). Setting this equal to zero tells us that there is a singular point at \(x = 0\).

• Singular points are important because they define where certain techniques or solutions may not apply effectively.
• General solutions often exclude or have special forms near singular points.

Identifying these points helps in understanding the limitations of solutions and ensures they are valid for a given range, excluding the singular points.

General Solution

The general solution of a differential equation is a formula or set of formulas that describes all possible solutions of the equation. It means finding a function \(y(x)\) that satisfies the differential equation for any given set of initial conditions. For the given equation, the general solution is derived using various techniques such as finding an integrating factor or changing variables. The constants of integration, \(C_1\) and \(C_2\), appear in the general solution to account for these initial conditions.

• The general solution helps predict how the system described by the differential equation will behave under different scenarios.
• Having a general solution is crucial as it can be adjusted to fit specific problems by applying initial or boundary conditions.

In our exercise, the solution must not include the singular point to remain valid.

Variable Coefficients

Differential equations with variable coefficients have coefficients for \(y''\), \(y'\), or \(y\) that are functions of \(x\), rather than constants. In the equation \(x^{2} y'' - 4 x y' + 4 y = 0\), \(x^2\) and \(-4x\) are variable coefficients. Having variable coefficients means the equation is more complex to solve compared to constant coefficients. This complexity arises from the changing behavior of the coefficients with \(x\).

• Solutions often involve techniques like substitution or integrating factors.
• Variable coefficients affect the nature of solutions, especially where they become zero (as in singular points).

Examples of such equations occur in real-life scenarios with changing conditions, like an electric circuit with varying resistance or a beam experiencing variable loads.