Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Term Test 1 => Topic started 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 ${y(\pi)=\pi, y'(\pi)=0}$.

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.
$y(t) = x + (1π) sinx$

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

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

$$
\text{emm just parts of solution？？}
$$
Thanks for waiting. Now I finish all solutions ;)

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

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