Question

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}-(x-0.1875) y=0, \quad x>0 ; \quad y_{1}(x)=x^{1 / 4} e^{2 \sqrt{x}}\)

Short answer

Based on the step by step solution, find a second solution to the given second-order linear differential equation using the method of reduction of order. The given differential equation is \[x^{2} y''(x) - (x-0.1875)y(x) =0\] with a known solution \(y_1(x) = x^{1/4} e^{2 \sqrt{x}}\). The method of reduction of order involves finding a new function \(v(x)\) such that \(y_2(x) = y_1(x)v(x)\). After following the steps in the provided solution, we end up solving a first-order linear homogeneous differential equation for \(u(x)\): \[u'(x) + 2\frac{y_1'(x)}{y_1(x)}u(x) = 0\] Find \(I(x)\), \(u(x)\), \(v(x)\), and finally \(y_2(x)\).

Step-by-step solution

Substitute the expression for \(y_2(x)\) into the differential equation

The proposed second solution is defined as \(y_2(x) = y_1(x) v(x)\), with known \(y_1(x) = x^{1/4} e^{2 \sqrt{x}}\). We will substitute this expression into the given differential equation and calculate the necessary derivatives in order to create the first-order linear differential equation for \(v(x)\). $$ y_2'(x) = y_1'(x)v(x) + y_1(x)v'(x), \\ y_2''(x) = y_1''(x)v(x) + 2y_1'(x)v'(x) + y_1(x)v''(x) $$

Simplify to get a first-order linear differential equation

Now we substitute the expressions for \(y_2\), \(y_2'\), and \(y_2''\) into the given differential equation: $$ x^{2}\left(y_1''(x)v(x) + 2y_1'(x)v'(x) + y_1(x)v''(x)\right) - (x-0.1875)\left(y_1(x)v(x)\right) = 0 $$ We can simplify this equation by noting that \(y_1(x)\) is a solution to the differential equation, so it satisfies: $$ x^{2}y_1''(x) - (x-0.1875)y_1(x)=0 $$ Thus, we can simplify the first-order equation for \(v(x)\): $$ x^{2} \left(2y_1'(x)v'(x) + y_1(x)v''(x)\right) = 0 $$ Dividing by \(x^2 y_1(x)\): $$ 2\frac{y_1'(x)}{y_1(x)}v'(x) + v''(x) = 0 $$ We have now reduced this problem to solving a first-order linear differential equation for \(v(x)\).

Solve the first-order linear differential equation

We have the first-order linear inhomogeneous differential equation: $$ v''(x) + 2\frac{y_1'(x)}{y_1(x)}v'(x) = 0 $$ At this point, let \(u(x) = v'(x)\), so \(v''(x) = u'(x)\). Then our differential equation becomes: $$ u'(x) + 2\frac{y_1'(x)}{y_1(x)}u(x) = 0 $$ This is now a first-order linear homogeneous differential equation, which can be solved using an integrating factor. The integrating factor is given by: $$ I(x) = e^{\int 2\frac{y_1'(x)}{y_1(x)} dx} $$ Compute \(I(x)\), multiply the differential equation by it, and solve for \(u(x)\) by integrating. Finally, find \(v(x)\) by integrating \(u(x)\) and then multiply \(y_1(x)\) by \(v(x)\) to get \(y_2(x)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that link a function with its derivatives. They play a crucial role in modeling various phenomena in engineering, physics, biology, and many other fields. In this context, they help us understand how the function is changing over time or space.

A differential equation can be ordinary or partial. **Ordinary differential equations** (ODEs), like the one in the exercise, involve functions of a single variable. They can be classified into different orders based on the highest derivative present. For example, our equation involves second derivatives, so it's considered a second-order ODE.

Solving differential equations often requires finding a function that satisfies the equation. Depending on the complexity, solutions might need to incorporate advanced techniques, such as reduction of order, which is what we are exploring here.

Second Solution

The concept of a second solution is pivotal in solving second-order differential equations. Often, a single solution is not sufficient to fully describe the system at hand, and that’s where finding a second solution comes in.

After identifying an initial solution, finding a second, linearly independent solution allows us to construct the general solution of the differential equation. This is especially vital when equations have variable coefficients.

In this exercise, we already have a known solution, denoted as \( y_1(x) \). The goal is to find a complementary solution, \( y_2(x) \), to form the most general solution, which is a linear combination of \( y_1(x) \) and \( y_2(x) \). This requires using methods like reduction of order to construct a valid second solution.

First-Order Linear Differential Equation

A first-order linear differential equation is an equation that involves the first derivative of a function and the function itself. These equations are fundamental in various fields as they describe processes that change at a rate proportional to the current state.

In the context of the exercise, after reducing the second-order differential equation by substituting the initial solution and simplifying, we end up with a first-order differential equation format in terms of \( v(x) \). These adjustments are made to simplify the problem, making it easier to solve by leveraging first-order methods.

This simplification results in an equation of the form:

\[ u'(x) + 2\frac{y_1'(x)}{y_1(x)}u(x) = 0 \]

Solving this simplified equation is a key step in eventually deriving the second solution.

Integrating Factor

The integrating factor is a powerful tool for solving linear first-order differential equations. It is a function that, when multiplied by the original differential equation, enables easy integration and solution extraction.

For our first-order simplified differential equation, we need to find an integrating factor \( I(x) \). This is calculated as:

\[ I(x) = e^{\int 2\frac{y_1'(x)}{y_1(x)} dx} \]

Once computed, the integral part in the exponent helps transform the differential equation into a form that can be neatly integrated. Multiplying the equation by the integrating factor allows us to express it as a derivative of a function, enabling straightforward integration to solve for \( u(x) \).

Through this process, we can find \( u(x) \) and subsequently integrate it to discover \( v(x) \), thus completing our second solution by using it alongside our original solution \( y_1(x) \).