Question

Find a power series solution in powers of x-1 of the initial-value problem.
xy''(x) + y'(x) + 2y(x) = 0, y(1) = 2, y'(1) = 4

Answer 1

Answer on Question #82286 – Math – Differential Equations

Question

Find a power series solution in powers of $x - 1$ of the initial-value problem.

$$x y ^ {\prime \prime} (x) + y ^ {\prime} (x) + 2 y (x) = 0, \qquad y (1) = 2, y ^ {\prime} (1) = 4$$

Solution

$$y (x) = y (1) + y ^ {\prime} (1) (x - 1) + \frac {y ^ {\prime \prime} (1)}{2 !} (x - 1) ^ {2} + \frac {y ^ {\prime \prime \prime} (1)}{3 !} (x - 1) ^ {3} + \frac {y ^ {4} (1)}{4 !} (x - 1) ^ {4} + \dots$$ $$y ^ {\prime \prime} (x) = - \frac {y ^ {\prime} (x) + 2 y (x)}{x}$$ $$y ^ {\prime \prime} (1) = - (4 + 2 \cdot 2) = - 8$$ $$y ^ {\prime \prime \prime} (x) = - \frac {x \left(y ^ {\prime \prime} (x) + 2 y ^ {\prime} (x)\right) - y ^ {\prime} (x) - 2 y (x)}{x ^ {2}}$$ $$y ^ {\prime \prime \prime} (1) = 8 - 2 \cdot 4 + 4 + 2 \cdot 2 = 8$$ $$y ^ {(4)} (x) = - \frac {x \left(y ^ {\prime \prime \prime} (x) + 2 y ^ {\prime \prime} (x)\right) - y ^ {\prime \prime} (x) - 2 y ^ {\prime} (x)}{x ^ {2}} + \frac {x ^ {2} \left(y ^ {\prime \prime} (x) + 2 y ^ {\prime} (x)\right) - 2 x \left(y ^ {\prime} (x) + 2 y (x)\right)}{x ^ {4}}$$ $$y ^ {(4)} (1) = - (8 - 2 \cdot 8) - 8 + 2 \cdot 4 - 8 + 2 \cdot 4 - 2 \cdot (4 + 2 \cdot 2) = - 8$$ $$y (x) = 2 + 4 (x - 1) - \frac {8}{2 !} (x - 1) ^ {2} + \frac {8}{3 !} (x - 1) ^ {3} - \frac {8}{4 !} (x - 1) ^ {4} + \dots$$ $$y (x) = 2 + 4 (x - 1) - 4 (x - 1) ^ {2} + \frac {4}{3} (x - 1) ^ {3} - \frac {1}{3} (x - 1) ^ {4} + \dots$$

Answer:

$$y (x) = 2 + 4 (x - 1) - 4 (x - 1) ^ {2} + \frac {4}{3} (x - 1) ^ {3} - \frac {1}{3} (x - 1) ^ {4} + \dots$$

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