Problem 20 - Differential Equations and Linear Algebra

Question

Determine two linearly independent solutions to the given differential equation on \((0, \infty)\) $$x^{2} y^{\prime \prime}-x^{2} y^{\prime}-2 y=0$$

Step-by-Step Solution

Answer

The differential equation is a second-order linear homogeneous equation with variable coefficients: \[y^{\prime \prime} - y^{\prime} -\frac{2}{x^{2}} y=0\] We performed the reduction of order, resulting in the equation: \[y'(x)= 2e^{x}\int\frac{e^{-x}}{x^2}y(x) dx + Ce^x\] After integrating once more, we obtained: \[y(x)= 2\int e^{x}\left[\int\frac{e^{-x}}{x^2}y(x) dx\right] dx + C_1e^x + C_2\] However, we did not find an explicit solution due to the equation having variable coefficients and requiring specialized methods.

Step 1: Identify the differential equation type and rewrite the equation

The given differential equation is a second-order linear homogeneous equation with variable coefficients: \[x^{2} y^{\prime \prime}-x^{2} y^{\prime}-2 y=0\] Let's rewrite the equation by dividing both sides by the leading coefficient, which is \(x^{2}\): \[y^{\prime \prime} - y^{\prime} -\frac{2}{x^{2}} y=0\]

Step 2: Perform the reduction of order

To perform the reduction of order, let's define a new function \(v(x)\) such that \(v(x) = y'(x)\). Then, the equation becomes a first-order linear equation for \(v(x)\): \[v'(x) - v(x) = \frac{2}{x^2} y(x)\]

Step 3: Solve the first-order linear equation for v(x) using an integrating factor

Let's introduce a new function, the integrating factor \(I(x)\), which is given by: \[I(x)=e^{\int{-1 dx}}=e^{-x}\] Now, we multiply the equation for \(v(x)\) by \(I(x)\) to obtain: \[e^{-x}v'(x) - e^{-x}v(x) = \frac{2}{x^2} y(x)e^{-x}\] The left-hand side of the equation is now an exact derivative: \[\frac{d}{dx} [ e^{-x}v(x)] = \frac{2}{x^2} y(x)e^{-x}\] Now, we integrate both sides of the equation with respect to \(x\): \[e^{-x}v(x) = 2\int\frac{e^{-x}}{x^2}y(x) dx + C\] Finally, we solve the equation for \(v(x)\): \[v(x) = 2e^{x}\int\frac{e^{-x}}{x^2}y(x) dx + Ce^x\]

Step 4: Recover the original equation

Now we can obtain the original equation by substituting back \(y'(x) = v(x)\), and then integrating with respect to \(x\): \[y'(x)= 2e^{x}\int\frac{e^{-x}}{x^2}y(x) dx + Ce^x\] Integrate both sides again with respect to \(x\): \[y(x)= 2\int e^{x}\left[\int\frac{e^{-x}}{x^2}y(x) dx\right] dx + C_1e^x + C_2\] At this point, we have two arbitrary constants \(C_1\) and \(C_2\), which correspond to two linearly independent solutions for the original equation. However, as the expression is still given in terms of the function \(y(x)\), we are unable to explicitly find the form of these two solutions. Note that using reduction of order on the given differential equation did not result in an explicitly solvable equation. In practice, it is often best to determine if there is a specific method, such as the Frobenius method or other techniques, tailored to the given equation type.

Key Concepts

Second-Order Differential Equation

In mathematics, a second-order differential equation is one in which the highest derivative of the unknown function is the second derivative. These equations are vital in modeling physical phenomena, such as the oscillation of springs, heat conduction, or wave propagation.

They are often written in the general form:

• \( a(x) y'' + b(x) y' + c(x) y = g(x) \)

The prime symbols signify differentiation with respect to the independent variable, usually time or space. When the function \(g(x)\) is zero, the equation is called homogeneous. If not, it's non-homogeneous.

In this case, we encountered a homogeneous equation:

• \( x^{2} y''-x^{2} y'-2 y=0 \)

Transforming the equation by dividing through by \(x^2\) simplifies the analysis and solution process, allowing us to use various methods to find solutions.

Reduction of Order

Reduction of order is a technique used to solve differential equations, usually when one solution to a second-order differential equation is already known. However, in some cases, it serves as a tool to simplify or solve new forms of equations.

The primary idea is to reduce the equation's order, making it more straightforward to deal with, particularly when faced with a complex or stubborn second-order differential equation.

• By assuming a function, say, \( v(x) \), is derivative of the unknown function \( y(x) \), the equation can often become manageable as a first-order equation.

In our specific example, we changed variables to \( v(x) = y'(x) \), simplifying the original second-order differential equation into a first-order form, ultimately aiding in applying other methods like integrating factors.

Integrating Factor

The integrating factor is a powerful method for solving linear first-order differential equations. It transforms an otherwise challenging differential equation into an easily integrable form.

Essentially, it is a function, often denoted \( I(x) \), that when multiplied by the original differential equation, facilitates transforming it into an exact differential equation. As a result, the equation can be rewritten as the derivative of a product.

The process follows these steps:

• Identify your linear equation in the form \( y' + p(x)y = q(x) \).
• Compute the integrating factor \( I(x) = e^{\int p(x) dx} \).
• Multiply every term by \( I(x) \), and observe that the left side of the equation becomes the derivative of \( I(x)y(x) \).
• Integrate both sides with respect to \( x \) and solve for \( y(x) \).

For our case, using the integrating factor \( I(x)=e^{-x} \), we transformed the modified equation for \( v(x) \), making it possible to express it as an integral, simplifying the path to the solution of the original problem.