The point $x = 0$ is a regular singular point of the given differential equation: $3x y'' - y' + 3y = 0$

• (a) Show that the indicial roots $r$ of the singularity do not differ by an integer. (List the indicial roots below as a comma-separated list.)

• (b) Use the method of Frobenius to obtain two linearly independent series solutions about $x = 0$. Form the general solution on $(0, \infty)$.

Neetesh Kumar | October 24, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh6Sec03 (Homework) Question - 4
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Step-by-step solution:

Part (a): Indicial Roots

The given differential equation is:

$3x y'' - y' + 3y = 0$

This second-order linear differential equation has a regular singular point at $x = 0$.
We will apply the Frobenius method to find the indicial equation and the roots.

Step 1: Rewrite the Equation in Standard Form

Divide the entire equation by $x$ to obtain the standard form:

$y'' - \frac{1}{3x} y' + \frac{1}{x}y = 0$

Now, compare this with the general form:

$y'' + \frac{P(x)}{x} y' + \frac{Q(x)}{x^2} y = 0$

From this, we identify:

• $P(x) = -\frac{1}{3}$
• $Q(x) = 1$

Step 2: Frobenius Method

Assume a solution of the form:

$y(x) = x^r \sum_{n=0}^{\infty} a_n x^n$

Substitute this into the differential equation and collect terms for the lowest power of $x$. The indicial equation comes from the coefficient of the lowest power of $x$.

Step 3: Indicial Equation

The indicial equation is obtained from the lowest power of $x$, typically involving $r(r-1)$ and terms with $r$. After solving, we get the roots of the indicial equation.

The indicial equation simplifies to:

$r(r - \frac{4}{3}) = 0$

Solving this gives the indicial roots:

$r = 0, \frac{4}{3}$

Thus, the indicial roots are:

$r = 0, \frac{4}{3}$

These roots do not differ by an integer.

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Part (b): Series Solution and General Solution

We now use the Frobenius method to find two linearly independent series solutions around $x = 0$.

Step 1: First Solution

We substitute into the series and solve for the first root $r_1 = 0$. The first solution has the form:

$y_1(x) = C_1 \left( 1 + 3x - \frac{9x^2}{4} + \frac{9x^3}{20} + \dots \right)$

Step 2: Second Solution

For the second root $r_2 = \frac{4}{3}$, the second linearly independent solution will have the form:

$y_2(x) = C_2 x^{4/3} \left( 1 - \frac{3x}{7} + \frac{9x^2}{140} - \frac{9x^3}{1820} + \dots \right)$

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Final Answer:

The general solution to the differential equation is a combination of the two independent solutions:

$y(x) = C_1 \left( 1 + 3x - \frac{9x^2}{4} + \frac{9x^3}{20} + \dots \right) + C_2 x^{4/3} \left( 1 - \frac{3x}{7} + \frac{9x^2}{140} - \frac{9x^3}{1820} + \dots \right)$

From the given options, this corresponds to the Last option:

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