Problem 2 (noon)

[Victor Ivrii]

(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE 
\begin{equation*}
\bigl(x\cos(x)-\sin(x)\bigr)y''+x\sin(x)y'-\sin(x)y=0.
\end{equation*}
(b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(\pi)=\pi, y'(\pi)=0}$.

[Nan Yang]

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.

$y(t) = x + (1-π) sinx$

[Yiheng Bian]

$$
\text{emm just parts of solution？？}
$$

[Yiheng Bian]

$$
\text{So when you have solve W then through}
$$
$$
 \left\{
 \begin{matrix}
   y_1 & y_2 \\
   y_1' & y_2'\\
  \end{matrix}
  \right\} \tag{2} \text{is equal to W solve y_2}
$$

[Nan Yang]

$$
\text{emm just parts of solution？？}
$$

Thanks for waiting. Now I finish all solutions 

[Yiheng Bian]

I think in part b when solve y_2 you skip many steps
the equation is
$$
xy_2'-y_2=xcosx-sinx
$$
$$
\int{xy_2'-y_2}=\int{xcos-sinx}
$$
So
$$
xy_2=xsinx
$$
Therefore
$$
y_2=sinx
$$

[Nan Yang]

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.