Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.) $(x - 1) y'' - x y' + y = 0$ with the initial conditions: $y(0) = -2, \, y'(0) = 5$

Neetesh Kumar | October 24, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh6Sec02 (Homework) Question - 5
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Step-by-step solution:

We are tasked with solving the given second-order differential equation using the power series method and satisfying the initial conditions.

Step 1: Assume a Power Series Solution

We assume the solution is of the form:

$y(x) = \displaystyle\sum_{n=1}^{\infty} a_n x^n$

Then, we have the following expressions for the derivatives of $y(x)$:

1. First derivative:

$y'(x) = \displaystyle\sum_{n=1}^{\infty} n a_n x^{n-1}$

2. Second derivative:

$y''(x) = \displaystyle\sum_{n=1}^{\infty} n(n-1) a_n x^{n-2}$

Step 2: Substitute into the Differential Equation

Now, substitute the power series expressions for $y(x)$, $y'(x)$, and $y''(x)$ into the differential equation:

$(x - 1) y''(x) - x y'(x) + y(x) = 0$

Substitute $y''(x)$, $y'(x)$, and $y(x)$ into the equation:

$(x - 1) \displaystyle\sum_{n=1}^{\infty} n(n-1) a_n x^{n-2} - x \sum_{n=1}^{\infty} n a_n x^{n-1} + \sum_{n=0}^{\infty} a_n x^n = 0$

We expand each term:

1. First term: $(x - 1) y''(x)$

$(x - 1) \displaystyle\sum_{n=1}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}$

2. Second term: $-x y'(x)$

$- x \displaystyle\sum_{n=1}^{\infty} n a_n x^{n-1} = - \sum_{n=1}^{\infty} n a_n x^n$

3. Third term: $y(x)$

$\displaystyle\sum_{n=1}^{\infty} a_n x^n$

Now combine the three terms:

$\displaystyle\sum_{n=1}^{\infty} n(n-1) a_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=0}^{\infty} a_n x^n = 0$

Substitute (n = (n+1)) in the first term and (n = (n + 2)) in the second term.

Rewriting the above expression again:

$\displaystyle\sum_{n=1}^{\infty} (n+1)(n) a_{n+1} x^{n} - \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n} - \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=0}^{\infty} a_n x^n = 0$

Now we have to make all summation terms starting from (n = 1)

$\displaystyle\sum_{n=1}^{\infty} (n+1)(n) a_{n+1} x^{n} - 2a_2 - \sum_{n=1}^{\infty} (n+2)(n+1) a_{n+2} x^{n} - \sum_{n=1}^{\infty} n a_n x^n + a_0 + \sum_{n=1}^{\infty} a_n x^n = 0$

Rearranging the above expression:

$(a_0 - 2a_2) + \displaystyle\sum_{n=1}^{\infty} (n+1)(n) a_{n+1} x^{n} - \sum_{n=1}^{\infty} (n+2)(n+1) a_{n+2} x^{n} - \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=1}^{\infty} a_n x^n = 0$

After solving further:

$(a_0 - 2a_2) + \displaystyle\sum_{n=1}^{\infty} \bigg((n^2+n)a_{n+1} - (n^2+3n+2)a_{n+2} + (1-n)a_n\bigg)x^n = 0$

Comparing coefficients of constants terms:

$a_0 - 2a_2 = 0 \Rarr a_2 = \frac{a_0}{2}$

Comparing coefficients of $x^n$ term:

$a_{n+2} = \dfrac{(n^2+n)a_{n+1}-(n-1)a_n}{n^2+3n+2}$, where $n \ge 1$

For $n=1 \Rarr a_3 = \dfrac{a_2}{3} = \dfrac{a_0}{6} = \dfrac{a_0}{1.2.3}$

For $n=2 \Rarr a_4 = \dfrac{6a_3-a_2}{12} = \dfrac{a_0}{24} = \dfrac{a_0}{1.2.3.4}$

For $n=3 \Rarr a_5 = \dfrac{12a_4-2a_3}{20} = \dfrac{a_0}{120} = \dfrac{a_0}{1.2.3.4.5}$

WE know that

$y(x) = \displaystyle\sum_{n=1}^{\infty} a_n x^n = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5 + .... \infty$

Replacing the values of coefficients in terms of $a_0$

$y(x) = \displaystyle\sum_{n=1}^{\infty} a_n x^n = a_0 \bigg(1 + \frac{x^2}{1.2} + \frac{x^3}{1.2.3} + \frac{x^4}{1.2.3.4} + \frac{x^5}{1.2.3.4.5} + ....\infty \bigg) + a_1x$

$y(x) = \displaystyle\sum_{n=1}^{\infty} a_n x^n = a_0 \bigg(1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + ....\infty \bigg) + a_1x$

We know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ....$

$y(x) = a_0(e^x - x) + a_1x$

Step 3: Apply the Initial Conditions

Using the given initial conditions $y(0) = -2$ and $y'(0) = 5$, we know the following:

• From $y(0) = -2$: this implies that $a_0 = -2$.

$y'(x) = a_0(e^x - 1) + a_1$

• From $y'(0) = 5$: this implies that $a_1 = 5$.

Step 5: Summing the Power Series

$y(x) = -2(e^x - x) + 5x$

After simplifying

$y(x) = 7x -2e^x$

Final Answer:

The final solution, formatted as an elementary function, is:

$y(x) = \boxed{7x -2e^x}$

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