Toronto Math Forum

Thanks for your advise. For showing a solution, I need write more details.  BUT I think it is very clear to see that $y_2$ is $sin(x)$  and it can save time for us in writing a test.   

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.

$y(t) = x + (1-π) sinx$

we got the same question and I type the solution of this question.
Question: find the general solution of the given differential equation
y'' $-$ 2y' $-$ 2y = 0

Solution:
Let y'' $-$ 2y' $-$ 2y = 0 be equation (1)
we assume that $y= e^{rt}$ is solution of (1)
Then we have:
  $y= e^{rt}$ 
$y'= re^{rt}$
$y''= r^2e^{rt}$
we substitute them into equation (1),
we have $ r^2e^{rt}$ $-$ 2$re^{rt}$ $-$ 2$e^{rt}$ = 0 ,
$ e^{rt}$$(r^2 - 2r -2) = 0$
since $ e^{rt}$ is not zero
so $(r^2 - 2r -2) = 0$
we can get r = $\frac{2 \pm \sqrt{4 +8}}{2}$
simplify it we get  r = $\frac{2 \pm 2\sqrt{3}}{2}$
Then r = $1+ \sqrt{3}$ r = $1- \sqrt{3}$
Then two roots of equation (1) is  $e^{(1+ \sqrt{3})t}$ and $e^{(1- \sqrt{3})t}$
Therefore, the general solution is  y =$c_1$ $e^{(1+ \sqrt{3})t}$ + $c_2$ $e^{(1- \sqrt{3})t}$