Question

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Answer 1

ANSWER on Question #58667 - Math - Differential Equations

QUESTION

(25 marks) Find $\alpha > 0$ such that $y_{1}(x) = x^{\alpha}$ is a solution of the following differential equation:

$$y'' - \frac{y'}{x} + \frac{y}{x^{2}} = 0, \quad x \in (0; +\infty).$$

Solve the given differential equation.

SOLUTION

If $y_{1}(x) = x^{\alpha}$ is a solution of the differential equation (1), then

$$y_{1}'' - \frac{y_{1}'}{x} + \frac{y_{1}}{x^{2}} = 0,$$ $$\begin{array}{l} y_{1}'' - \frac{y_{1}'}{x} + \frac{y_{1}}{x^{2}} = \alpha (\alpha - 1) x^{\alpha - 2} - \frac{\alpha x^{\alpha - 1}}{x} + \frac{x^{\alpha}}{x^{2}} \\ = \alpha (\alpha - 1) x^{\alpha - 2} - \alpha x^{\alpha - 2} + x^{\alpha - 2} = 0 \\ \end{array}$$ $$\begin{array}{l} (\alpha (\alpha - 1) - \alpha + 1) x^{\alpha - 2} = 0 \rightarrow \alpha (\alpha - 1) - \alpha + 1 = 0 \\ \alpha (\alpha - 1) - \alpha + 1 = \alpha^{2} - \alpha - \alpha + 1 = \alpha^{2} - 2\alpha + 1 \\ = (\alpha - 1)^{2} = 0 \rightarrow \alpha = 1 \\ \end{array}$$

$y_{1}(x) = x$ is the first solution of the differential equation (1).

It follows from Liouville's formula that the second solution is given by

$$\begin{array}{l} y_{2}(x) = y_{1} \int \frac{e^{-\int p(x) dx}}{y_{1}^{2}} dx = x \int \frac{e^{-\int_{-1}^{x} dx}}{x^{2}} dx = x \int \frac{e^{\ln(x)}}{x^{2}} dx = x \int \frac{x}{x^{2}} dx = \\ = x \int \frac{1}{x} dx = x \ln(x) \\ \end{array}$$

The general solution of the differential equation (1) is

$$y(x) = C_{1} y_{1}(x) + C_{2} y_{2}(x) = C_{1} x + C_{2} x \ln(x)$$

Answer: $\alpha = 1$; $y(x) = C_1 x + C_2 x \ln(x)$.

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