Question

Use the Frobenius method to obtain one solution of the following ODE:
4 yx ′′ + 2y′ + y = 0

Answer 1

Answer on Question #66346, Physics / Mechanics | Relativity

Use the Frobenius method to obtain one solution of the following ODE: $4 \, \text{yx}'' + 2 \, \text{y}' + \text{y} = 0$

Solution:

$$a _ {k + 1} = - a _ {k} / (2 k + 2 r + 2) (2 k + 2 r + 1), k = 0, 1, 2 \dots (1)$$

To find $y_1$ apply (14) with $r = r_1 = 1 / 2$ to get the recurrence relation

$$a _ {k + 1} = - a _ {k} / (2 k + 3) (2 k + 2), k = 0, 1, 2 \dots$$

Then

$$a _ {1} = - a _ {0} / 3 \cdot 2, a _ {2} = - a _ {1} / 5 \cdot 4, a _ {3} = - a _ {2} / 7 \cdot 6$$ $$\text {S o} a _ {1} = - a _ {0} / 3!, a _ {2} = a _ {0} / 5!, a _ {3} = - a _ {0} / 7!$$

Since $a_0$ is arbitrary, let $a_0 = 1$

$$a _ {k} \left(r _ {1}\right) = (- 1) ^ {k} / (2 k + 1)!, k = 0, 1, 2 \dots$$ $$y _ {1} (x) = x ^ {1 / 2} \sum_ {k = 0} ^ {\infty} \left(\left(- 1\right) ^ {k} / (2 k + 1)!\right) x ^ {k}$$

To find $y_2$ , just apply (1) with $r = r_2 = 0$ to get the recurrence relation

$$b _ {k + 1} = - b ^ {k} / (2 k + 2) (2 k + 1)$$

Letting the arbitrary constant $b_0 = 1$ , then

$$b _ {k} \left(r _ {2}\right) = (- 1) ^ {k} / (2 k)$$ $$y _ {2} (x) = \sum_ {k = 0} ^ {\infty} \left(\left(- 1\right) ^ {k} / (2 k)\right) x ^ {k}$$

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