Question

Use the power series method to obtain one solution of the following ODE:
3 0
2
x y ′′ + y′ − xy

Answer 1

Answer on Question #62066 – Math – Differential Equations

Question

Use the power series method to obtain one solution of the following ODE:

$$xy'' + y' - xy = 0,$$ $$x_0 = 2, \, y(x_0) = y(2) = 3, \, y'(x_0) = y'(2) = 0.$$

Solution

$$y(x_0) = 3, \, y'(x_0) = 0, \, x_0 = 2$$ $$xy'' + y' - xy = 0$$ $$y'' = \frac{xy - y'}{x} = y - \frac{y'}{x}$$ $$y''(2) = y(2) - \frac{y'(2)}{2} = 3 - 0 = 3$$ $$y''' = \left(y - \frac{y'}{x}\right)' = y' - \left(\frac{y''}{x} - \frac{y'}{x^2}\right) = y' - \frac{y''}{x} + \frac{y'}{x^2}$$ $$y'''(2) = y'(2) - \frac{y''(2)}{2} + \frac{y'(2)}{4} = -\frac{3}{2}$$ $$y(x) = y(x_0) + \frac{y'(x_0)}{1!}(x - x_0) + \frac{y''(x_0)}{2!}(x - x_0)^2 + \frac{y'''(x_0)}{3!}(x - x_0)^3 + \dots = 3 + \frac{3}{2}(x - 2)^2 - \frac{3}{12}(x - 2)^3 + \dots$$

Answer: $y(x) = 3 + \frac{3}{2}(x - 2)^2 - \frac{3}{12}(x - 2)^3 + \dots$

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