### Author Topic: Q3: TUT0102  (Read 901 times)

#### Gavrilo Milanov Dzombeta

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##### Q3: TUT0102
« on: October 11, 2019, 02:00:06 PM »
$$\text{Verify that the functions } y_1 \text{ and } y_2 \text{ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?}$$
$$\left( 1- x \cot(x)\right) y^{\prime \prime} - x y^{\prime} + y = 0,$$
$$0 < x < \pi \text{ and } y_1 (x) = x \text{ and } y_2 (x) = \sin(x)$$

$$\text{Consider } y_1(x) = x$$
$$\therefore {y_1}^{\prime}(x) = 1 \text{ and } {y_1}^{\prime \prime}(x) = 0$$
$$(1 - x\cot(x))y^{\prime \prime} - xy^{\prime} + y = (1 - x\cot(x))(0) - x + x = 0$$
$$\therefore y_1(x) = x \text{ is a solution of the differential equation } (1 - x\cot(x))y^{\prime \prime} - x y^{\prime} + y = 0$$
$$\text{Consider } y_2(x) = \sin(x)$$
$$y^{\prime} = \cos(x) \text{ and } y^{\prime \prime} = \sin(x)$$
$$(1 - x\cot(x))y^{\prime \prime} - x y^{\prime} + y = (1 - x\cot(x))(-\sin(x)) - x(\cos(x)) + \sin(x) = 0$$
$$W(y_1, y_2)(t) = \begin{array}{|cc|} x & \sin(x) \\ 1 & \cos(x) \end{array} = x\cos(x) - \sin(x) \ne 0 \text{ for } 0 < x < \pi$$
$$\therefore y_1, y_2 \text{ form a fundamental set of solutions}$$