Solution

**a) **

$y'' + \frac{xsin(x)}{xcos(x)-sin(x)}y'- \frac{sin(x)}{xcos(x)-sin(x)}y = 0$

$p(x) = \frac{xsin(x)}{xcos(x)-sin(x)}$

$w = ce^{-\int p(x)dx} =ce^{-\int \frac{xsin(x)}{xcos(x)-sin(x)}} $

let $u = xcos(x)-sin(x), du =-xsin(x) $

$w = ce^{- \int \frac{-1}{u} du} = ce^{ \int \frac{1}{u} du} = ce^{lnu} = ce^{ln(xcos(x)-sin(x))} = c(xcos(x)-sin(x))$

let $ c = 1 , w = xcos(x)-sin(x)$

**b)**

check $y_1 =x$ is a solution.

$y_1' = 1, y_2'' = 0$

substitute them into equation,

we get

$xsin(x)- sin(x)x = 0$

so x is a solution

w = $\begin{vmatrix}

x& y_2 \\

1 & y_2'

\end{vmatrix}$

$xy_2' - y_2 = xcos(x)-sin(x) $

so $y_2 = sinx$ OK. V.I.

**c)**

$y(t) = c_1x + c_2 sinx$

since $y(π) = π, y'(π) = 0$

$π = c_1 π + c_2sin(π)$

$π = c_1π $

so $c_1 = 1$

$y'(t) = c_1 + c_2 cos(x)$

$π = 1 - c2$

$c_2 = 1 -π $ Wrong.

$y(t) = x + (1-π) sinx$