Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started by: Gavrilo Milanov Dzombeta on October 11, 2019, 02:00:06 PM

$$ \text{Verify that the functions } y_1 \text{ and } y_2 \text{ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?} $$
$$\left( 1 x \cot(x)\right) y^{\prime \prime}  x y^{\prime} + y = 0,$$
$$0 < x < \pi \text{ and } y_1 (x) = x \text{ and } y_2 (x) = \sin(x) $$
$$ $$
$$ \text{Consider } y_1(x) = x $$
$$ \therefore {y_1}^{\prime}(x) = 1 \text{ and } {y_1}^{\prime \prime}(x) = 0$$
$$ (1  x\cot(x))y^{\prime \prime}  xy^{\prime} + y = (1  x\cot(x))(0)  x + x = 0$$
$$ \therefore y_1(x) = x \text{ is a solution of the differential equation } (1  x\cot(x))y^{\prime \prime}  x y^{\prime} + y = 0$$
$$ \text{Consider } y_2(x) = \sin(x) $$
$$ y^{\prime} = \cos(x) \text{ and } y^{\prime \prime} = \sin(x) $$
$$ (1  x\cot(x))y^{\prime \prime}  x y^{\prime} + y = (1  x\cot(x))(\sin(x))  x(\cos(x)) + \sin(x) = 0 $$
$$W(y_1, y_2)(t) = \begin{array}{cc} x & \sin(x) \\ 1 & \cos(x) \end{array} = x\cos(x)  \sin(x) \ne 0 \text{ for } 0 < x < \pi $$
$$\therefore y_1, y_2 \text{ form a fundamental set of solutions}$$