Question

Use the method of reduction of order to find a second solution of the given differential equation. \((x-1) y^{\prime \prime}-x y^{\prime}+y=0, \quad x>1 ; \quad y_{1}(x)=e^{x}\)

Short answer

Question: Use the method of reduction of order to find the second linearly independent solution, \(y_2(x)\), given the differential equation \((x-1)y''-xy'+y=0\) and \(y_1(x) = e^x\). Answer: \(y_2(x) = (\ln|x-1|)e^x\)

Step-by-step solution

Differentiate \(y_1(x)\) and \(y_2(x)\)

We first need to find the derivatives of \(y_1(x) = e^x\) and \(y_2(x) = u(x)e^x\). For \(y_1(x)\), we have: \(y_1'(x) = \frac{d}{dx}(e^{x}) = e^{x}\) \(y_1''(x) = \frac{d^2}{dx^2}(e^{x}) = e^{x}\) For \(y_2(x)\): \(y_2'(x) = \frac{d}{dx}(u(x)e^{x}) = u'(x)e^{x} + u(x)e^{x}\) \(y_2''(x) = \frac{d^2}{dx^2}(u(x)e^{x}) = u''(x)e^x + 2u'(x)e^x + u(x)e^{x}\) Now that we have the derivatives, we can substitute them into the original equation.

Substitute \(y_2(x)\) and its derivatives

Replace \(y\), \(y'\), and \(y''\) with the expressions we found for \(y_2(x)\), \(y_2'(x)\) and \(y_2''(x)\): \((x-1)(u''(x)e^x + 2u'(x)e^x + u(x)e^{x}) - x(u'(x)e^{x} + u(x)e^{x}) + u(x)e^{x} = 0\)

Simplify the equation

Factor out \(e^{x}\) and arrange the terms: \(e^{x}[(x-1)(u''(x) + 2u'(x) + u(x)) - xu'(x) - xu(x) + u(x)] = 0\)

Solve for \(u'(x)\)

Since \(e^x\) is never equal to zero, we can divide the whole equation by \(e^x\) without changing the value of the equation. After simplification and regrouping the terms, we get: \((x-1)u''(x) + u'(x) = 0\) To make it easier to solve, use the substitution \(v(x) = u'(x)\). Thus, \(v'(x) = u''(x)\) and the equation becomes: \((x-1)v'(x) + v(x) = 0\) Now integrate both sides of the expression with respect to \(x\): \(\int \frac{dv}{dx}(x-1)dx = -\int dx\)

Solve for \(v(x)\) and \(u(x)\)

Upon integrating both sides, we obtain: \(v(x) = \frac{C}{x-1}\) Now remember that \(v(x) = u'(x)\), so to find \(u(x)\), integrate \(v(x)\) with respect to \(x\): \(u(x) = \int \frac{C}{x-1}dx = C\ln|x-1| + D\)

Find \(y_2(x)\)

Recall that \(y_2(x) = u(x)y_1(x)\), so substitute the found \(u(x)\) and the given \(y_1(x)\) into this expression: \(y_2(x) = (C\ln|x-1| + D)e^x\) Since we are interested in the linearly independent solution only, we can disregard the constant \(D\), which would be absorbed into \(y_1(x)\), thus the second linearly independent solution is: \(y_2(x) = (\ln|x-1|)e^x\)

Key concepts

Second Order Differential Equations

Second order differential equations involve derivatives up to the second degree – essentially, they include the second derivative of a function. The general form is usually \[ a(x)\frac{d^2y}{dx^2} + b(x)\frac{dy}{dx} + c(x)y = g(x), \] where \(a(x), b(x), c(x)\) are coefficients that may depend on \(x\), and \(g(x)\) is a given function. These equations are common in physics and engineering, describing phenomena such as motion and forces.

There are various methods to solve second order differential equations that depend on their characteristics. In homogeneous equations where \(g(x) = 0\), solutions often take a simpler form as they rely only on the structure of the differential equation. By finding two linearly independent solutions, we can express the general solution of the equation.

Linear Independence

In the context of differential equations, linear independence is a crucial concept. Two functions \( y_1(x) \) and \( y_2(x) \) are linearly independent if no constant \( c \) exists for which \[ y_2(x) = c \cdot y_1(x) \] for all \(x\) in their domain. This means that one solution cannot be formed by merely scaling the other.

For second order differential equations, having two linearly independent solutions allows us to find the general solution, which is a linear combination of these solutions. Such a combination enables capturing all possible behaviors of the system described by the differential equation. The method of reduction of order often helps in finding a second linearly independent solution when one solution is already known.

Method of Integration

The method of integration in solving differential equations often involves finding antiderivatives to determine the function \( u(x) \) whose derivatives fulfill the criteria set by the differential equation. When using the reduction of order method, after substituting a presumed form for the solution into the original differential equation, integration becomes crucial to solve for unknown functions.

For instance, given the differential equation after substitution and simplification, we perform integration to deduce subsequent expressions until obtaining \( u(x) \). This requires integrating terms step by step:

• First, you integrate to find \( v(x) = u'(x) \).
• Then, integrating \( v(x) \) to finally derive the function \( u(x) \).
• This cumulative process essentially builds up from simpler derivatives until the solution is found.

Integration, therefore, acts as a tool that transforms differential parts into the cumulative function that satisfies the entire differential equation.

Differential Equation Solving Techniques

Differential equation solving techniques are diverse, decided based on the type and complexity of the equation at hand. The reduction of order method is a strategic technique applicable for non-homogeneous second order linear differential equations when one solution is already known.

This approach substitutes a function that involves the known solution to reset the problem into a solvable form, usually easier to manage. One assumes a solution of the form \( y = u(x)y_1(x) \), where \( y_1(x) \) is the known solution:

• Differentiate this expression to integrate into the original equation.
• Simplify and factor out dependencies, often resulting in a simpler first order equation.
• Upon solving, employ integration to recover the form of \( u(x) \).

This ingenious repackaging of the problem not only helps in solving challenging second order equations but also deepens the understanding of solution spaces and their properties.