Problem 15 - Advanced Engineering Mathematics

Question

\(x=0\) is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about \(x=0\). Form the general solution on the interval \((0, \infty)\). $$ 2 x y^{\prime \prime}-y^{\prime}+2 y=0 $$

Step-by-Step Solution

Answer

The indicial roots are \(0\) and \(\frac{1}{2}\), and the general solution is a combination of the two series solutions.

Step 1: Write the Differential Equation in Standard Form

First, ensure your differential equation is in the form required for applying the method of Frobenius. The standard form for this type of solution is \[ x^2 y'' + x P(x) y' + Q(x) y = 0 \]The given equation is \(2x y'' - y' + 2y = 0\). Divide the entire equation by 2 to rewrite it:\[ x y'' - \frac{1}{2} y' + y = 0 \] Now it's in the appropriate form with \(P(x) = -\frac{1}{2x}\) and \(Q(x) = \frac{1}{x}\).

Step 2: Determine the Indicial Equation

The indicial equation arises from substituting a series solution into the differential equation and finding the lowest power contribution when \(x \to 0\). Suppose a solution of the form \[ y = x^r \sum_{n=0}^{fty} a_n x^n \]Insert this into the equation and evaluate the terms involving the lowest power of \(x\) for the leading coefficient:\[ O(x^{r-1}): \, r(2r-1)a_0 = 0 \] Thus, the indicial equation is \[ r(2r-1) = 0 \] Giving roots \(r = 0\) and \(r = \frac{1}{2}\). Since the roots do not differ by an integer, we can proceed with two independent solutions given by the method of Frobenius.

Step 3: Calculate the First Solution for r = 0

For the solution corresponding to \(r = 0\), assume \[ y_1 = \sum_{n=0}^{fty} a_n x^n \]Substitute this series into the differential equation and equate like powers of \(x\). Solving the recurrence relation provides the coefficients:- Assuming \(a_0 eq 0\), all other coefficients are determined.The recurrence gives a series solution for \(r = 0\).

Step 4: Calculate the Second Solution for r = \(\frac{1}{2}\)

For the solution corresponding to \(r = \frac{1}{2}\), assume \[ y_2 = x^{\frac{1}{2}} \sum_{n=0}^{fty} b_n x^n \] Substitute back into the differential equation, similar to the previous step, and solve the recurrence relation.- With \(b_0 eq 0\), you will again solve for each \(b_n\).Develop the second series solution for \(r = \frac{1}{2}\).

Step 5: Form the General Solution on (0, ∞)

The general solution of the differential equation is a linear combination of the two linearly independent series solutions obtained:\[ y(x) = C_1 y_1(x) + C_2 y_2(x) \]where \(C_1\) and \(C_2\) are arbitrary constants.Thus, you have expressed the complete general solution and verified that it covers the interval \((0, \infty)\).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. These equations describe various physical phenomena, such as heat, light, sound, elasticity, and fluid dynamics. In simple terms, they tell us how a particular quantity changes over time or space. In our exercise, the given differential equation is:\[2 x y^{\prime \prime} - y^{\prime} + 2 y = 0\]Rewriting it in standard form, as required for the Frobenius method, gives us:\[x y^{\prime \prime} - \frac{1}{2} y^{\prime} + y = 0\]This form is necessary for identifying various coefficients that will allow us to apply the method of series solutions. By tackling such equations, we can serve several applications to predict and understand complex systems and their behaviors.

Indicial Equation

An indicial equation helps determine possible values (also known as roots) for the exponent when we apply series solutions, especially near singular points. For Frobenius' method, it particularly identifies characteristics of these roots. When we write the solution as:\[y = x^r \sum_{n=0}^{\infty} a_n x^n\]and substitute it into the differential equation, the indicial equation emerges to look at terms of the lowest power of \(x\). In our case, we derived:\[r(2r - 1) = 0\]Solving this provides roots: \(r = 0\) and \(r = \frac{1}{2}\). These roots guide how we form our general solution, indicating they are not different by an integer, fulfilling a primary requirement to proceed with the Frobenius method.

Regular Singular Point

A regular singular point in a differential equation is where the standard form of the equation exhibits specific behavior but remains manageable using series solutions. Regular singular points are crucial because they provide the right conditions for the Frobenius method to work. Our differential equation has a regular singular point at \(x = 0\). This stems from the equation's behavior:- The function \(P(x)\) has a singularity weaker than an essential singularity since it behaves as \(-\frac{1}{2x}\).- The function \(Q(x)\) also illustrates limited behavior as \(\frac{1}{x}\).Understanding these functions' behaviors ensures that the Frobenius method can generate solutions that won't deviate into "rough" mathematical territory. At such points, rightful use of series solutions provides leverage to handle the equation elegantly.

Series Solutions

Series solutions involve expressing the variable of interest in the differential equation as a sum of terms, often involving powers of \(x\). Such solutions are typically sought around singular points. By assuming a solution of the form \(y = \sum_{n=0}^{\infty} a_n x^n\), one can determine the coefficients \(a_n\) using the recurrence relation derived from the differential equation.For our exercise, we developed two series solutions corresponding to the different indicial roots:- **For \(r = 0\):** We assume \(y_1 = \sum_{n=0}^{\infty} a_n x^n\). This yields the first series solution, where all coefficients are systematically determined under consistency.- **For \(r = \frac{1}{2}\):** We take \(y_2 = x^{\frac{1}{2}} \sum_{n=0}^{\infty} b_n x^n\). Solving for each \(b_n\) gives the second series solution.Finally, the general solution combines these independent series solutions:\[y(x) = C_1 y_1(x) + C_2 y_2(x)\]where \(C_1\) and \(C_2\) are arbitrary constants, providing a comprehensive solution over the interval \((0, \infty)\).