Problem 16 - Advanced Engineering Mathematics

Question

In Problems \(1-16\), the indicated function \(y_{1}(x)\) is a solution of the given 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 \(y_2(x) = -\ln|1-x^2| + C_2\).

Step 1: Differentiation of the first solution

Start with the given solution \(y_1(x) = 1\). Compute its first derivative: \(y_1'(x) = 0\) and its second derivative: \(y_1''(x) = 0\).

Step 2: Prepare the Reduction of Order Formula

To find the second solution using the reduction of order, assume a solution of the form \(y_2(x) = v(x)y_1(x)\). Since \(y_1(x)=1\), this reduces to \(y_2(x) = v(x)\). The derivatives become \(y_2'(x) = v'(x)\) and \(y_2''(x) = v''(x)\).

Step 3: Substitute into the Differential Equation

Substitute \(y = v(x)\), \(y' = v'(x)\), and \(y'' = v''(x)\) into the original differential equation: \[(1-x^2)v''(x) + 2xv'(x) = 0\].

Step 4: Solve the Resulting Equation

Recognize the form of the equation: \[v''(x) + \frac{2x}{1-x^2}v'(x) = 0\] is a first-order equation in terms of \(v'(x)\). Let \(w = v'(x)\), then \(w'(x) + \frac{2x}{1-x^2}w = 0\). This can be separated and solved by integration.

Step 5: Integrate to find \(w(x)\)

Separate variables: \[\frac{dw}{w} = -\frac{2x}{1-x^2} dx\]Integrate both sides: \[\ln|w| = \ln(1-x^2) + C_1\]Solve for \(w\): \[ w(x) = \frac{C_1}{1-x^2}\]

Step 6: Integrate to find \(v(x)\)

Recall that \(w(x) = v'(x)\). So, integrate again:\[ v(x) = \int \frac{C_1}{1-x^2} \, dx = -C_1 \ln|1-x^2| + C_2\]Assume \(C_1 = 1\) for simplicity, then:\[ v(x) = -\ln|1-x^2| + C_2\].

Step 7: Express the Second Solution

Therefore, the second solution is:\[ y_2(x) = v(x) = -\ln|1-x^2| + C_2 \].

Key Concepts

Reduction of Order

Reduction of order is a technique used in solving linear homogeneous differential equations when one solution is already known. It allows us to find a second linearly independent solution. This is crucial because, for second-order differential equations, having two solutions enables us to form the general solution. In this method, we assume that the second solution can be written in the form \(y_2(x) = v(x)y_1(x)\), where \(v(x)\) is a function to be determined. Given that the known solution \(y_1(x) = 1\), the equation simplifies as \(y_2(x) = v(x)\). By substituting \(y_2(x)\) and its derivatives into the differential equation, we reduce the problem to solving for \(v(x)\). This approach is straightforward when one solution is already given and helps to uncover another solution that complements it to solve the differential equation entirely.

Second Solution

Finding a second solution for a differential equation is essential in constructing the general solution. By finding a linearly independent solution to the already known one, we ensure that we can describe the entire solution space of the differential equation. In our exercise, the second solution \(y_2(x)\) was found using the reduction of order technique. We expressed \(y_2(x)\) as \(v(x)\), since \(y_1(x) = 1\). The process to find this second solution involved steps like setting \(v'(x)\) to \(w(x)\) and transforming the differential equation appropriately. Through integration, after changing the variable, we arrived at the expression \(y_2(x) = -\ln|1-x^2| + C_2\). This shows how the integration essentially helped "build up" the second solution from the simpler structure provided by \(y_1\). It's a logical way to approach differential equations when initial solutions provide a path.

Homogeneous Differential Equation

Homogeneous differential equations are a class of differential equations that play a crucial role in mathematical modeling where functions and their derivatives relate in a balance. These equations have the property that if \(y = f(x)\) is a solution, then \(y = cf(x)\) is also a solution for any constant \(c\). Thus, they maintain "scalability" of solutions. For this reason, solutions must be linearly independent, which means no solution is a constant multiple of another. This forms the basis for needing additional solutions beyond the first one. In our problem, the differential equation \((1-x^{2})y'' + 2xy' = 0\) is homogeneous because each term involves either the function \(y\) or its derivatives, with no independent terms involved. Solving such equations typically involves finding multiple solutions, such as our \(y_1\) and \(y_2\), to form the general solution, which is critical for modeling and prediction.