Question

(a) Verify that \(y_{1}=1 /(x-1)\) and \(y_{2}=1 /(x+1)\) are solutions of $$\left(x^{2}-1\right) y^{\prime \prime}+4 x y^{\prime}+2 y=0$$ on \((-\infty,-1),(-1,1),\) and \((1, \infty) .\) What is the general solution of \((\mathrm{A})\) on each of these intervals? (b) Solve the initial value problem $$\left(x^{2}-1\right) y^{\prime \prime}+4 x y^{\prime}+2 y=0, \quad y(0)=-5, \quad y^{\prime}(0)=1$$ What is the interval of validity of the solution? (c) \(\mathrm{C} / \mathrm{G}\) Graph the solution of the initial value problem. (d) Verify Abel's formula for \(y_{1}\) and \(y_{2}\), with \(x_{0}=0\).

Short answer

In summary: a) The given possible solutions \(y_1 = \frac{1}{x-1}\) and \(y_2 = \frac{1}{x+1}\) are indeed valid solutions to the given differential equation. The general solution on each interval is \(y(x) = C_{1}\frac{1}{x-1} + C_{2}\frac{1}{x+1}\). b) The particular solution for the given initial value problem is \(y(x) = -3\frac{1}{x-1} -2\frac{1}{x+1}\), and its interval of validity is \((-1, 1)\). c) Graph the solution function \(y(x) = -3\frac{1}{x-1} -2\frac{1}{x+1}\) for \(x\in(-1,1)\). {{graph image of this function}} d) Abel's formula is satisfied for \(y_1\) and \(y_2\) with \(x_0=0\) when \(c=-1\).

Step-by-step solution

Verify if given functions are valid solutions

Using the given functions \(y_{1}\) and \(y_{2}\), we must first find their first and second derivatives: $$y_{1} = \frac{1}{x - 1}, \quad y_{1}^{\prime} = -\frac{1}{(x - 1)^2}, \quad y_{1}^{\prime\prime} = \frac{2}{(x - 1)^3}$$ $$y_{2} = \frac{1}{x + 1}, \quad y_{2}^{\prime} = -\frac{1}{(x + 1)^2}, \quad y_{2}^{\prime\prime} = \frac{2}{(x + 1)^3}$$ Now, substitute those expressions back into the differential equation: $$\left(x^{2}-1\right) \left( \frac{2}{(x - 1)^3} \right)+4 x \left( -\frac{1}{(x - 1)^2} \right)+2 \left( \frac{1}{x - 1} \right) = 0$$ $$\left(x^{2}-1\right) \left( \frac{2}{(x + 1)^3} \right)+4 x \left( -\frac{1}{(x + 1)^2} \right)+2 \left( \frac{1}{x + 1} \right) = 0$$ After simplifying the expressions, you will find that the differential equation is satisfied by both \(y_{1}\) and \(y_{2}\). They are indeed valid solutions on the given intervals.

Find the general solution

Now, we can construct the general solution by using a linear combination of \(y_{1}\) and \(y_{2}\): $$y(x) = C_{1}y_{1} + C_{2}y_{2} = C_{1}\frac{1}{x-1} + C_{2}\frac{1}{x+1}$$ This is the general solution on each of the three intervals: \((-\infty, -1)\), \((-1,1)\), and \((1, \infty)\). ##Part (b): Solve the initial value problem and find the interval of validity##

Find constants

We are given the initial conditions \(y(0) = -5\) and \(y'(0) = 1\). We need to find the constants \(C_{1}\) and \(C_{2}\) in our general solution that satisfy these conditions. $$-5 = C_{1}\frac{1}{-1} + C_{2}\frac{1}{1}$$ $$1 = -C_{1}\frac{2}{1} - C_{2}\frac{2}{1}$$ Solving these simultaneous equations, we find that \(C_{1} = -3\) and \(C_{2}=-2\). Now we can write the particular solution for these initial conditions: $$y(x) = -3\frac{1}{x-1} -2\frac{1}{x+1}$$

Determine the interval of validity

We have discontinuities at \(x = -1\) and \(x = 1\), so the interval of validity for the solution of the initial value problem will be \((-1, 1)\), where the solution is continuous and differentiable. ##Part (c): Graph the solution of the initial value problem## Plot the function \(y(x) = -3\frac{1}{x-1} -2\frac{1}{x+1}\) for \(x\in(-1,1)\). {{graph image of this function}} ##Part (d): Verify Abel's formula for \(y_{1}\) and \(y_{2}\), with \(x_{0}=0\).## Abel's formula is given by: $$y_{1}(x)y_{2}^{\prime}(x)-y_{1}^{\prime}(x)y_{2}(x) = c(y_{1}(x_{0})y_{2}^{\prime}(x_{0})-y_{1}^{\prime}(x_{0})y_{2}(x_{0}))$$ We have already calculated the derivatives of the two functions: $$y_{1}^{\prime} = -\frac{1}{(x - 1)^2} \quad \text{and} \quad y_{2}^{\prime} = -\frac{1}{(x + 1)^2}$$ We will now plug in the values of \(x_0 = 0\) into Abel's formula and check if \(y_1\) and \(y_2\) satisfy the formula: $$\left(\frac{1}{1}\right)\left(-\frac{1}{1}\right) - \left(-\frac{1}{1}\right)\left(\frac{1}{1}\right) = c\left(\frac{1}{1}\left(-\frac{1}{1}\right) - \left(-\frac{1}{1}\right)\left(\frac{1}{1}\right)\right)$$ This reduces to \(-1 - 1 = 2c\). We can see that for \(c=-1\), Abel's formula is satisfied for \(y_1\) and \(y_2\) with \(x_0=0\).

Key concepts

Initial Value Problem

An initial value problem (IVP) in the context of differential equations involves finding a specific solution to a differential equation that satisfies given initial conditions at a certain point. This type of problem is crucial because differential equations often have infinitely many solutions, and the initial conditions allow us to pinpoint a unique solution that describes a particular physical scenario or experimental data.

For the exercise provided, the IVP is formulated with an initial value for the function and its first derivative, namely, \( y(0) = -5 \) and \( y'(0) = 1 \). The goal is to determine the constants in the general solution that will satisfy these initial conditions. This process typically involves plugging the initial values into the general solution and its derivative to form a system of equations, which we solve to find the particular constants specific to the IVP. Once the constants are determined, we obtain a particular solution that adheres to the given constraints.

General Solution of Differential Equations

The general solution of a differential equation encompasses all possible solutions to the equation and usually includes arbitrary constants. For second-order linear differential equations, as seen in the provided exercise, the general solution can be a linear combination of two linearly independent solutions. In simpler terms, if you find two different solutions that work for the equation, generally any weighted sum of those two will also be a solution.

In our case, \( y_1 = \frac{1}{x-1} \) and \( y_2 = \frac{1}{x+1} \) are both solutions to the given differential equation. The general solution is then expressed as \( y(x) = C_1y_1 + C_2y_2 \), where \( C_1 \) and \( C_2 \) are constants that can be tailored to fit specific initial conditions, thus transforming the general solution into a particular solution for a given IVP.

Abel's Formula

Abel's formula, also known as Abel's identity, is a useful tool in the theory of second-order linear differential equations. It provides a relationship between two solutions to a homogeneous differential equation. Essentially, it tells us that the Wronskian of two solutions (a determinant involving the solutions and their derivatives) is proportional to the exponential of the opposite of an integral involving the coefficients of the equation.

In our exercise, Abel's formula is used to verify that the two solutions \( y_1 \) and \( y_2 \) are indeed linearly independent and satisfy the differential equation on the specified intervals. By applying Abel's formula with \( x_0 = 0 \), we can validate the consistency of the provided solutions with the differential equation. It's a powerful check that confirms we're on the right track with our solutions.

Interval of Validity

The interval of validity for a solution to a differential equation is the range over which the solution is well-defined and satisfies the equation. This concept is crucial because, in real-world scenarios, not every solution can be applied universally due to the presence of singularities or the specific domain of the problem at hand.

In the initial value problem from the exercise, the particular solution has discontinuities where the denominator of the solution becomes zero, which happens at \( x=-1 \) and \( x=1 \). Therefore, even though we found a solution that works on paper, it's only valid within the interval \( -1 < x < 1 \), because outside of this interval, the solution would involve division by zeros and thus would become undefined. This interval represents the domain within which the solution is continuous and differentiable, adhering to the initial conditions specified in the IVP.