Question

Use the method of Problem 33 to find a second independent solution of the given equation. \((x-1) y^{\prime \prime}-x y^{\prime}+y=0, \quad x>1 ; \quad y_{1}(x)=e^{x}\)

Short answer

In this exercise, we were given a linear 2nd-order inhomogeneous differential equation: $$(x-1) y^{\prime \prime}-x y^{\prime}+y=0, \quad x>1$$ and a known independent solution, \(y_1(x) = e^x\). We used the method of Reduction of Order to find a second independent solution. We expressed the second independent solution in the form \(y_2(x) = v(x) y_1(x) = v(x)e^{x}\), and proceeded to substitute it into the given equation, which eventually resulted in a first-order linear homogeneous differential equation. We solved this equation by introducing an integrating factor to obtain \(v'(x) = \frac{C e^x}{x^2}\). Then, we integrated \(v'(x)\) to find \(v(x)\) in terms of the integral: $$v(x) = C \int \frac{e^x}{x^2} dx + C_\text{1}$$ Finally, we expressed the second independent solution as: $$y_2(x) = \left(C \int \frac{e^x}{x^2} dx + C_\text{1} \right)e^x$$ where \(C\) and \(C_\text{1}\) are arbitrary constants.

Step-by-step solution

Write Down the Given Equation and Known Solution

We are given the following 2nd-order linear inhomogeneous differential equation and a known solution: $$(x-1) y^{\prime \prime}-x y^{\prime}+y=0, \quad x>1 ; \quad y_1(x)=e^{x}$$

Finding the second solution of the form \(y_2(x) = v(x) y_1(x)\)

We are going to find a second independent solution in the form: $$y_2(x) = v(x) y_1(x) = v(x)e^{x}$$

Calculate the Derivatives of \(y_2(x)\)

To substitute the form of \(y_2(x)\) into the given equation, we need its first and second derivatives. First derivative: $$y_2^{\prime}(x) = v(x)e^{x} + v'(x)e^{x} = e^x (v(x) + v'(x))$$ Second derivative: $$y_2^{\prime \prime}(x) = e^x (v(x) + 2v'(x) + v''(x))$$

Substitute \(y_2(x)\), \(y_2^{\prime}(x)\), and \(y_2^{\prime \prime}(x)\) into the Given Equation and Simplify

Substitute \(y_2(x)\), \(y_2^{\prime}(x)\), and \(y_2^{\prime \prime}(x)\) into the given equation: $$(x-1) e^x (v(x) + 2v'(x) + v''(x)) - x e^x (v(x) + v'(x)) + e^x v(x) = 0$$ Dividing by \(e^x\): $$(x-1)(v(x) + 2v'(x) + v''(x)) - x(v(x) + v'(x)) + v(x) = 0$$ Now, simplify the equation to get: $$xv''(x) + (2-x)v'(x) = 0$$

Solve for \(v'(x)\)

Divide both sides by \(x\) and rewrite the equation as: $$v''(x) + \frac{(2-x)}{x}v'(x) = 0$$ This is a first-order linear homogeneous differential equation. We can first find the integrating factor, \(R(x)\), where: $$R(x) = e^{\int \frac{(2-x)}{x} dx}$$ Calculating the integral: $$\int \frac{(2-x)}{x} dx = \int \frac{2}{x} dx - \int dx = 2\ln(x) - x$$ Now compute the integrating factor: $$R(x) = e^{2\ln(x) - x} = x^2e^{-x}$$

Multiply the Differential Equation by the Integrating Factor and Solve for \(v'(x)\)

Multiply both sides by the integrating factor, \(x^2e^{-x}\): $$x^2e^{-x} v''(x) + (2-x)e^{-x} x v'(x) = 0$$ Notice that the left side is the derivative of \(x^2e^{-x} v'(x)\) with respect to x. Therefore, integrate the equation with respect to x. $$\int\left[x^2e^{-x} v''(x) + (2-x)e^{-x} x v'(x)\right] dx = \int 0 dx$$ The left side becomes: $$x^2e^{-x} v'(x) = C$$ Solve for \(v'(x)\): $$v'(x) = \frac{C e^x}{x^2} $$

Solve for \(v(x)\)

Now, integrate \(v'(x)\) to find \(v(x)\): $$v(x) = \int v'(x) dx = \int \frac{C e^x}{x^2} dx + C_\text{1}$$ This integral does not have an elementary solution. However, we can write \(v(x)\) in terms of the integral: $$v(x) = C \int \frac{e^x}{x^2} dx + C_\text{1}$$

Write the Second Independent Solution$

Finally, we can now obtain \(y_2(x)\) as a function of the integral: $$y_2(x) = v(x) y_1(x) = \left(C \int \frac{e^x}{x^2} dx + C_\text{1} \right)e^x$$ where \(C\) and \(C_\text{1}\) are arbitrary constants.

Key concepts

Second Order Differential Equations

Second order differential equations involve the second derivative of a function. These equations are frequently used in physics and engineering to model systems that change over time, like oscillating springs or electrical circuits.

The general form of a second order differential equation is \(a(x)y'' + b(x)y' + c(x)y = g(x)\). Here, \(y''\) represents the second derivative, \(y'\) the first derivative, and \(y\) the function itself. The coefficients \(a(x)\), \(b(x)\), and \(c(x)\) are functions of \(x\), while \(g(x)\) is a given function or zero for homogeneous equations.

Solving these equations involves finding a function \(y(x)\) that satisfies the given equation. The challenge increases with the complexity of \(a(x), b(x), c(x)\), and \(g(x)\).

• Homogeneous second order: If \(g(x) = 0\), the equation is homogeneous.
• Inhomogeneous second order: Here \(g(x) eq 0\), which often requires more advanced methods to solve.

Linear Homogeneous Differential Equations

Linear homogeneous differential equations are a special category where the function \(g(x)\) equals zero, meaning the equation looks like \(a(x)y'' + b(x)y' + c(x)y = 0\).

These equations are fundamental because they tend to have properties and solutions that can be expressed in standard forms. The solutions are often expressed as a linear combination of independent solutions. Specifically, for second-order equations, there will generally be two linearly independent solutions that constitute the general solution.

The initial steps to solving them include:

• Finding a known solution, if given, or exploring special functions like exponentials or trigonometric functions.
• Utilizing methods such as reduction of order or variation of parameters to find a second solution, assuming one solution is known.
• Expressing the general solution as \(y(x) = C_1y_1(x) + C_2y_2(x)\), where \(C_1\) and \(C_2\) are arbitrary constants.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique to solve linear differential equations, specifically inhomogeneous ones, where the right-hand side consists of easily handled functions like polynomials, exponentials, or sines/cosines.

This approach works best when the solution can be guessed based on the form of the inhomogeneous part. For instance, if \(g(x)\) is a polynomial, an exponential, or a trigonometric function, a solution of similar form can be assumed.

The steps involve:

• Assuming a trial solution \(y_p(x)\) whose form mirrors the non-homogeneous term \(g(x)\).
• Plugging \(y_p(x)\) into the original differential equation and determining the coefficients to satisfy the equation.
• Adding this particular solution to the general solution of the homogeneous equation to derive the full solution.

One limitation is it doesn't work for every equation and depends on the simplicity of \(g(x)\). It's more suited for constant coefficient differential equations.

Integrating Factor Method

The integrating factor method is especially helpful for making some first-order linear differential equations more manageable, but it has adaptations for some second-order cases like those reduced to first order.

The basic idea is to multiply the differential equation by an integrating factor, which turns it into an exact equation or one that can be easily integrated.

For the equation \(v''(x) + p(x)v'(x) = 0\), the steps include:

• Determining an integrating factor \(R(x)\) by solving \(e^{\int p(x) dx}\).
• Multiplying through by \(R(x)\) to simplify the equation.
• Integrating the left-hand side directly, since it becomes the derivative of the product \(R(x) v'(x)\).
• Solving for \(v'(x)\) by integrating once more to find \(v(x)\), which is then used in building the solution.

This method shows how sometimes a more straightforward, direct integration can unriddle complex differential equations by strategic manipulation.