Problem 17 - Differential Equations and Linear Algebra

Question

Show that the indicial equation of the given differential equation has distinct roots that do not differ by an integer and find two linearly independent Frobenius series solutions on \((0, \infty)\). $$3 x^{2} y^{\prime \prime}+x(7+3 x) y^{\prime}+(1+6 x) y=0$$

Step-by-Step Solution

Answer

The indicial equation for the given differential equation is \(\rho^2 - \rho + \frac{7}{3} = 0\), with roots \(\rho_1, \rho_2 = \frac{1 \pm \sqrt{1- \frac{28}{3}}}{2}\), which are distinct and do not differ by an integer. The two linearly independent Frobenius series solutions on \((0, \infty)\) are \(y_{\rho_1}(x) = \sum_{k=0}^{\infty} C_{k,1} x^{k+\rho_1}\) and \(y_{\rho_2}(x) = \sum_{k=0}^{\infty} C_{k,2} x^{k+\rho_2}\), where the coefficients \(C_{k,1}\) and \(C_{k,2}\) are determined by the recurrent relation \[ C_k=\frac{-C_{k-1}-6C_{k-1}}{(\rho+k)(\rho+k-1)+\frac{7}{3}}\ ,\text{ for } k > 0. \]

Step 1: Simplify the given differential equation

Divide the given differential equation by \(3x^{2}\) to simplify it: \[\frac{3x^2y''}{3x^2} + \frac{x(7+3x)y'}{3x^2} + \frac{(1+6x)y}{3x^2} = \frac{0}{3x^2}\] This simplifies to: \[y'' + \frac{(7+3x)}{3x}y' + \frac{(1+6x)}{3x^2}y = 0\]

Step 2: Write the Frobenius series solution form and plug it into the simplified DE

We assume a Frobenius series solution of the form: \[ y_\rho(x) = \sum_{k=0}^{\infty} C_k x^{k+\rho} \] Then we calculate the first derivative \(y'_\rho(x)\) and second derivative \(y''_\rho(x)\): \[ y'_\rho(x) = \sum_{k=0}^{\infty} (k+\rho) C_k x^{k+\rho-1} \] \[ y''_\rho(x) = \sum_{k=0}^{\infty} (k+\rho)(k+\rho-1) C_k x^{k+\rho-2} \] Now, we substitute \(y_\rho(x)\), \(y'_\rho(x)\), and \(y''_\rho(x)\) into the simplified differential equation obtained in step 1: \[\sum_{k=0}^{\infty} (k+\rho)(k+\rho-1) C_k x^{k+\rho-2} + \frac{(7+3x)}{3x} \sum_{k=0}^{\infty} (k+\rho) C_k x^{k+\rho-1} + \frac{(1+6x)}{3x^2} \sum_{k=0}^{\infty} C_k x^{k+\rho} = 0\]

Step 3: Multiply each power series by the corresponding coefficient terms

Multiple each power series by the corresponding coefficient terms and group the terms by powers of \(x\), then set the coefficients of each power to 0: \[ \sum_{k=0}^{\infty} \left[ (k+\rho)(k+\rho-1) C_k + \frac{7+3x}{3x} (k+\rho)C_k x + \frac{1+6x}{3x^2}C_k x^2 \right] x^{k+\rho-2} = 0 \] We collect the terms with the same power of \(x\) to form equations: \[ x^{\rho-2} \left[ (\rho)(\rho-1)C_0 + \frac{7}{3}C_0 \right] + \] \[ x^{\rho-1} \left[ (\rho+1)(\rho)C_1 + \frac{7}{3}(\rho+1)C_1 + C_0 + 6C_0\right] + \] \[ x^\rho \left[ (\rho+2)(\rho+1)C_2 + \frac{7}{3}(\rho+2)C_2 + C_1 + 6C_1 \right] + \] \[ \cdots = 0 \] Now, we set each coefficient term to 0, which results in a system of recurrent relations and the indicial equation:

Step 4: Form the indicial equation and find its roots

Set the first coefficient term to 0: \[ (\rho)(\rho-1)C_0 + \frac{7}{3}C_0 = 0. \] This is the indicial equation. We can simplify it, notice that if \(C_0 \neq 0\), then \(\rho(\rho-1) + \frac{7}{3} = 0\). Now solve this quadratic equation for \(\rho\): \[ \rho^2 - \rho + \frac{7}{3} = 0 \] Using the quadratic formula, we have: \[ \rho_1, \rho_2 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 (1) \cdot \frac{7}{3}}}{2(1)} , \] which simplifies to: \[ \rho_1, \rho_2 = \frac{1 \pm \sqrt{1- \frac{28}{3}}}{2}. \] These roots are distinct, and they do not differ by an integer, so we can proceed with finding two linearly independent Frobenius series solutions.

Step 5: Use the recurrent relations to find the coefficients for the Frobenius series solutions

Using the recurrence relations obtained in Step 3: For \(k = 0\): \[ (\rho+1)(\rho)C_1 + \frac{7}{3}(\rho+1)C_1 + C_0 + 6C_0 = 0 \] We obtain the recurrent relation for the coefficients: \[ C_0 = C_0 \quad \text{for each } \rho \] \[ C_k=\frac{-C_{k-1}-6C_{k-1}}{(\rho+k)(\rho+k-1)+\frac{7}{3}}\ ,\text{ for } k > 0 \] Now, by setting \(C_0 = C_{0,1}\), and plugging in \(\rho = \rho_1\), we can compute the Frobenius series coefficients for the first solution \(y_{\rho_1}(x)\) and similarly for the second solution \(y_{\rho_2}(x)\) by setting \(C_0 = C_{0,2}\) and plugging in \(\rho = \rho_2\). Finally, we obtain the two linearly independent Frobenius series solutions on (0, \(\infty\)): \[ y_{\rho_1}(x) = \sum_{k=0}^{\infty} C_{k,1} x^{k+\rho_1} \] \[ y_{\rho_2}(x) = \sum_{k=0}^{\infty} C_{k,2} x^{k+\rho_2} \] since \(\rho_1\) and \(\rho_2\) do not differ by an integer, these two series represent linearly independent solutions of the given differential equation.

Key Concepts

Indicial Equation

The indicial equation plays a crucial role when we're trying to solve certain types of differential equations using Frobenius method. It arises when we plug a power series solution into a differential equation at a singular point, typically at the origin.

The indicial equation helps us determine the possible values for the exponent \rho in our Frobenius series. This is the starting exponent of the series, and it significantly shapes the form of the solution. To find the indicial equation, we examine the coefficient of the lowest power of \(x\) after substituting the Frobenius series.

In our problem, the indicial equation is \(\rho(\rho-1) + \frac{7}{3}\rho = 0\), leading to the calculation of roots that are essential for building the series solutions. If these roots are distinct and not differing by an integer, as they are in this case, we can guarantee that the solutions derived from these \/rho values are linearly independent, which means each solution provides unique information about the behavior of the differential equation's solutions near the singular point.

Differential Equations

Differential equations are mathematical equations that describe the relationship between functions and their derivatives. They are foundational in expressing physical laws where change is continuous, such as motion, growth, or decay.

The equation given, \(3 x^{2} y^{\'\'}+x(7+3 x) y^{\' }+(1+6 x) y=0\), is a second-order linear homogeneous differential equation. 'Homogeneous' means all terms involve the unknown function \(y\) or its derivatives; 'linear' signifies that the equation contains no powers or products of \(y\) and its derivatives each term is either a constant times \(y\) or a derivative of \(y\).

Solving these equations often requires special techniques. The Frobenius method, which expands on the power series method by allowing non-integer exponents in the series, is particularly useful for tackling differential equations with singular points - locations where the functions or their derivatives may become infinite or undetermined.

Linearly Independent Solutions

Typically, a second-order differential equation has two solutions. For a complete solution, these should be linearly independent, meaning that one cannot be expressed as a constant multiple or a linear combination of the other.

In our context, linear independence is essential to cover the full spectrum of behaviors that satisfy the differential equation under study. For the Frobenius series solutions to be useful, each series must bring unique traits to the table. The absence of linear dependence is why we need distinct roots for the indicial equation that do not differ by an integer.

Once we confirm this condition, as we did with the roots of our indicial equation, we can use the recurrence relations to find coefficients for our series. The two distinct Frobenius series solutions stemming from distinct roots are guaranteed to be linearly independent, offering a complete picture of the solution space for the differential equation.