Q15 E

Question

In Problems 13-19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem.

$(x^{2} + 1) y'' - e^{x} y' + y = 0; y (0) = 1, y' (0) = 1$

Step-by-Step Solution

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Answer

The first four nonzero terms in the power series expansion of the given initial value problem $(x^{2} + 1) y'' - e^{x} y' + y = 0$ is $y (t) = 1 + x + \frac{x^{4}}{24} + \frac{x^{5}}{60}$ .

1Step 1: Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

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

2Step 2: Find the relation.

Given,

$(x^{2} + 1) y'' - e^{x} y' + y = 0; y (0) = 1, y' (0) = 1$

Write the given equation in standard form

$y'' - \frac{e^{x}}{x^{2} + 1} y' + \frac{1}{x^{2} + 1} y = 0$

From the above equation, we get,

$p (x) = \frac{e^{x}}{x^{2} + 1}, q (x) = \frac{1}{x^{2} + 1}$

Use the formula,

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

Taking derivative and substituting in the equation, we get the relation,

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

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

Hence, we get the relation $(x^{2} + 1) \sum_{n = 2}^{\infty} n (n - 1) \cdot a_{n} {(x)}^{n - 2} - (e^{x}) \sum_{n = 1}^{\infty} n \cdot a_{n} {(x)}^{n - 1} + \sum_{n = 0}^{\infty} a_{n} {(x)}^{n} = 0$ .

3Step 3: Find the expression after expansion.

The series expansion for the function is

$(x^{2} + 1) (2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + 20 a_{5} x^{3} + 30 a_{6} x^{4} + \dots) - (1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots) \cdot (a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} + 5 a_{5} x^{4} + \dots) + (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots) = 0$

By expanding the series we get,

$(2 a_{2} + 6 a_{3} x + 12 a_{4} x^{2} + 20 a_{5} x^{3} + 30 a_{6} x^{4} + \dots) + (2 a_{2} x^{2} + 6 a_{3} x^{3} + 12 a_{4} x^{4} + 20 a_{5} x^{5} + 30 a_{6} x^{6} + \dots) - (a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} + 5 a_{5} x^{4} + \dots) - (a_{1} x + 2 a_{2} x^{2} + 3 a_{3} x^{3} + 4 a_{4} x^{4} + 5 a_{5} x^{5} \dots) - (a_{1} \frac{x^{2}}{2} + 2 a_{2} \frac{x^{3}}{2} + 3 a_{3} \frac{x^{4}}{2} + 4 a_{4} \frac{x^{5}}{2} + 5 a_{5} \frac{x^{6}}{2} + \dots) - (a_{1} \frac{x^{3}}{6} + 2 a_{2} \frac{x^{4}}{6} + 3 a_{3} \frac{x^{5}}{6} + 4 a_{4} \frac{x^{6}}{6} + 5 a_{5} \frac{x^{7}}{6} + \dots)$

Simplify the expression.

$(2 a_{2} + a_{0} - a_{1}) + (6 a_{3} + a_{1} - a_{1} - 2 a_{2}) x + (12 a_{4} + 2 a_{2} - 3 a_{3} - 2 a_{2} - \frac{a_{1}}{2} + a_{2}) x^{2} + (20 a_{5} + 6 a_{3} - 4 a_{4} - 3 a_{3} - \frac{2 a_{2}}{2} - \frac{a_{1}}{6} + a_{3}) x^{3} + \dots = 0$

Hence the expression after the expansion is:

4Step 4: Find the first four nonzero terms.

By equating the coefficients, we get,

$2 a_{2} + a_{0} - a_{1} = 0 \to a_{2} = 0 6 a_{3} + a_{1} - a_{1} - 2 a_{2} = 0 \to a_{3} = 0 12 a_{4} + 2 a_{2} - 3 a_{3} - 2 a_{2} - \frac{a_{1}}{2} + a_{2} = 0 \to a_{4} = \frac{1}{24} 20 a_{5} + 6 a_{3} - 4 a_{4} - 3 a_{3} - \frac{2 a_{2}}{2} - \frac{a_{1}}{6} + a_{3} \to a_{5} = \frac{1}{60}$