# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 1 => Topic started by: Victor Ivrii on October 23, 2019, 06:02:44 AM

Title: Problem 2 (noon)
Post by: Victor Ivrii on October 23, 2019, 06:02:44 AM
(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 ${y(\pi)=\pi, y'(\pi)=0}$.
Title: Re: Problem 2 (noon)
Post by: Nan Yang on October 23, 2019, 08:05:41 AM
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.

$y(t) = x + (1-π) sinx$
Title: Re: Problem 2 (noon)
Post by: Yiheng Bian on October 23, 2019, 08:19:56 AM
$$\text{emm just parts of solution？？}$$
Title: Re: Problem 2 (noon)
Post by: Yiheng Bian on October 23, 2019, 08:25:22 AM
$$\text{So when you have solve W then through}$$
$$\left\{ \begin{matrix} y_1 & y_2 \\ y_1' & y_2'\\ \end{matrix} \right\} \tag{2} \text{is equal to W solve y_2}$$

Title: Re: Problem 2 (noon)
Post by: Nan Yang on October 23, 2019, 08:51:47 AM
$$\text{emm just parts of solution？？}$$
Thanks for waiting. Now I finish all solutions  ;)
Title: Re: Problem 2 (noon)
Post by: Yiheng Bian on October 23, 2019, 09:06:44 AM
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
$$y_2=sinx$$
Title: Re: Problem 2 (noon)
Post by: Nan Yang on October 23, 2019, 12:08:59 PM
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.   ;D