The point $x = 0$ is a regular singular point of the given differential equation: $xy'' + 2y' - xy = 0$

• (a) Show that the indicial roots $r$ of the singularity differ by an integer. (List the indicial roots below as a comma-separated list.)

• (b) Use the method of Frobenius to obtain at least one series solution about $x = 0$. Use (23) in Section 6.3:

$y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$

where necessary, and use a CAS if instructed to find a second solution. 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 - 5
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Step-by-step solution:

Part (a): Indicial Roots

The given differential equation is:

$x y'' + 2y' - x y = 0$

This is a second-order linear differential equation with 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:

$y'' + \frac{2}{x} y' - y = 0$

Now, compare this with the general form:

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

From this, we identify:

• $P(x) = 2$
• $Q(x) = -1$

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 $x$.

After substituting into the equation and simplifying, we obtain the indicial equation:

$r(r - 1) + 2r = 0$

Simplifying this equation:

$r^2 + r = 0$

Factoring:

$r(r + 1) = 0$

The roots of the indicial equation are:

$r = 0, -1$

These roots differ by 1, confirming that the roots differ by an integer.

Thus, the indicial roots are:

$r = 0, -1$

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Part (b): Series Solution and General Solution

We now use the Frobenius method to find the general solution. We have two roots: $r_1 = 0$ and $r_2 = -1$. The solution for $r_1 = 0$ gives us the first solution, and we will use the integral formula to find the second solution for $r_2 = -1$.

Step 1: First Solution $y_1(x)$

For $r_1 = 0$, we substitute into the series solution and obtain:

$y_1(x) = 1$

Thus, the first solution is:

$y_1(x) = 1$

Step 2: Second Solution $y_2(x)$

To find the second solution $y_2(x)$, we use the formula:

$y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$

From earlier, $P(x) = 2$. So:

$\int P(x) dx = 2 \ln(x) = \ln(x^2)$

Therefore:

$e^{-\int P(x) dx} = e^{-\ln(x^2)} = \frac{1}{x^2}$

Now, substitute into the formula for $y_2(x)$:

$y_2(x) = 1 \cdot \int \frac{1}{x^2} dx = -\frac{1}{x}$

Thus, the second solution is:

$y_2(x) = -\frac{1}{x}$

Step 3: General Solution

The general solution is a linear combination of $y_1(x)$ and $y_2(x)$:

$y(x) = C_1 y_1(x) + C_2 y_2(x)$

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Step 4: Form of the Solution

Given the options provided in the problem, the form of the second solution and general solution fits with the third option, which is:

$y = \frac{1}{x} \left[C_1 \sinh x + C_2 \cosh x \right]$

This option represents the general solution regarding $\sinh$ and $\cosh$ functions.

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Final Answer:

• The indicial roots are $r = 0, -1$.
• The correct form of the solution is the third option:

$y(x) = \frac{1}{x} \left[C_1 \sinh x + C_2 \cosh x \right]$

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