Problem 16 - A First Course in Differential Equations with Modeling Applications

Question

The indicated function \(y_{1}(x)\) is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to find a second solution \(y_{2}(x)\). $$\left(1-x^{2}\right) y^{\prime \prime}+2 x y^{\prime}=0 ; \quad y_{1}=1$$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right| \).

Step 1: Understand the Problem

We are given a second-order homogeneous linear differential equation: \((1-x^2) y'' + 2xy' = 0 \), with one solution \( y_1(x) = 1 \). Our goal is to find a second, linearly independent solution \( y_2(x) \) using the reduction of order method.

Step 2: Use Reduction of Order Formula

The reduction of order method suggests using a second solution of the form \( y_2(x) = v(x)\cdot y_1(x) \), where \( v(x) \) is an unknown function to be determined. Substituting \( y_1(x) = 1 \) gives \( y_2(x) = v(x) \). So, we assume \( y_2(x) = v(x) \).

Step 3: Derive First and Second Derivatives

Since \( y_2(x) = v(x) \), calculate its derivatives: \( y_2' = v' \) and \( y_2'' = v'' \). Substitute these into the original differential equation.

Step 4: Substitute into the Differential Equation

Substitute \( y_2' = v' \) and \( y_2'' = v'' \) into the differential equation: \((1-x^2) v'' + 2x v' = 0.\)

Step 5: Simplify the Equation

This simplifies to the equation: \(v'' + \frac{2x}{1-x^2} v' = 0.\) This is a first order differential equation in \( v' \).

Step 6: Integrate the Equation

Recognize that this is a first order linear differential equation for \( v' \). Use separation of variables: \(dv' = -\frac{2x}{1-x^2} v' dx\). Integrate both sides to find \( v' \):\(\ln |v'| = -\ln |1-x^2| + C\) \(v' = \frac{K}{1-x^2}\) , where \( K = e^C \).

Step 7: Find Function v(x)

Integrate \( v' = \frac{K}{1-x^2} \):\( v = K \int \frac{1}{1-x^2} dx \) The integral \( \int \frac{1}{1-x^2} dx \) is the inverse hyperbolic tangent, \( v = \frac{K}{2}\ln\left|\frac{1+x}{1-x}\right| + C_1 \).

Step 8: Determine y2(x)

Using \( v(x) \), the second solution is \( y_2(x) = v(x)\cdot 1 = \frac{K}{2}\ln\left|\frac{1+x}{1-x}\right| + C_1 \). Because a constant solution does not affect linear independence, take \( K = 1 \) and \( C_1 = 0 \) (any multiple other than zero will suffice for independence): \( y_2(x) = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right| \).

Key Concepts

Second-Order Differential Equation

A second-order differential equation is a type of equation that involves the second derivative of a function. It's called "second-order" because the highest derivative is the second one:

• This means our function or dependent variable is differentiated twice.
• Second-order differential equations are common in physics and engineering because they can describe how systems evolve over time, such as oscillations or acceleration.

In the exercise problem, the equation is \((1-x^2) y'' + 2xy' = 0\), where \( y'' \) is the second derivative of the function \( y \). Notice that it involves both \( y' \), the first derivative, and \( y'' \), the second derivative.
Solving second-order differential equations often involves finding two solutions because solutions to these equations form a two-dimensional vector space.

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a specific type of differential equation that has solutions with a particular structure.

• Homogeneity means that the equation equals zero: \( L[y] = 0 \), where \( L \) is a linear differential operator.
• Linearity implies that if two functions are solutions, any linear combination of them is also a solution.

This characteristic allows us to solve the equation by breaking it into simpler parts. In our exercise, \((1-x^2) y'' + 2xy' = 0\) is homogeneous because the right-hand side is zero.
When a known solution is given (like \( y_1 = 1 \) in this problem), it helps us determine another solution using methods like reduction of order.

Linearly Independent Solution

Linearly independent solutions are fundamental to forming the general solution of differential equations.

• Two functions, \( y_1(x) \) and \( y_2(x) \), are linearly independent if no constant \( c \) exists such that \( y_2(x) = c \, y_1(x) \).
• Linear independence guarantees that the two solutions describe different aspects or modes of the system's behavior.

Having two linearly independent solutions allows us to express any solution to a homogeneous differential equation as a combination of those two. In the exercise, finding \( y_2(x) \) such that it is independent of \( y_1(x) = 1 \), is critical to formulating the general solution. For our equation, using the reduction of order technique helps us derive this independent second solution.

Hyperbolic Functions

Hyperbolic functions are analogs to trigonometric functions but for a hyperbola. They are encountered in solutions to differential equations involving the shape or behavior similar to hyperbolas.

• Common hyperbolic functions include \( \sinh(x) \), \( \cosh(x) \), \( \tanh(x) \).
• Their derivatives have similar properties to trigonometric functions, making them useful in calculus.

In our exercise, while solving the integral for \( v(x) \), we encounter the inverse hyperbolic tangent, \( \ln\left|\frac{1+x}{1-x}\right| \), which appears as part of the second solution \( y_2(x) \).
Hyperbolic functions often arise in solutions when dealing with equations that have hyperbolic symmetry or when integrating certain types of rational expressions.