Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:34:59 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*}
y''\sin ^2(x)y' \tan(x)(2+\cos^2(x))+3y=0.
\end{equation*}
(b) Check that $y_1(x)=\sin(x)$ is a solution and find another linearly independent solution.
(c) Write the general solution. Find solution such that ${y(\frac{\pi}{3})=0, y'(\frac{\pi}{3})=7}$.

(a) Rewrite the equation: $$y''  \frac{2+\cos^{2}x}{\sin x\cos x}y' + \frac{3}{sin^{2}x}y = 0.$$
Then $$p(x) =  \frac{2+\cos^{2}x}{\sin x\cos x} = \frac{2}{\sin x\ cos x}  \frac{\cos x}{\sin x}.$$
By Abel's Theorem, we have:
$$
W(y_1,y_2)(x) =c\exp \bigl(\int p(x)dx\bigr) = c\exp\bigl(\int(\frac{2}{\sin x \cos x} + \frac{\cos x}{sin x}) dx\bigr)=\\
C\exp \bigl(\int \frac{2}{\sin x\cos x} dx + \int \frac{\cos x}{\sin x} dx\bigr)$$
Since $\int( \frac{2}{sinxcosx} )dx = 2\int(\frac{1}{sinxcosx})dx = 2\int(\frac{sex}{sinx})dx$
$= 2\int\frac{sec^{2}x}{sinxsecx}dx = 2\int\frac{du}{u} =2lnu = 2ln(tanx) $
(by substitute: $u = tanx, du = sec^{2}xdx)$
and $\int(\frac{cosx}{sinx})dx = \int\frac{du}{u} = lnu = ln(sinx) $
(by substitute: $u = sinx, du = cosxdx$)
So, $$W(y_1,y_2)(x) = ce^{2\ln(\tan x) +\ln(\sin x)}= \tan^{2}x\sin x$$
(b) Since $y_1(x) = sinx$, so $y_1'(x) = cosx, y_1''(x) = sinx$
Plug in: $sinxsin^{2}x  tanx(2 + cos^{2}x)cosx + 3sinx$
= $sin^{3}x sinx(2 + 1  sin^{2}x) + 3sinx$
= $sin^{3}x 3sinx +sin^{3}x + 3sinx$
= 0
So, $y_1(x) = \sinx$ is a solution.
Take c = 1, then $W(y_1,y_2)(x) = tan^{2}xsinx$
By Reduction of Order, we have:
$y_2 = y_1\int(\frac{(e^{\intp(x)dx})}{y_1^{2}})dx$ = $sinx\int(\frac{tan^{2}xsinx}{sin^{2}x})dx$
= $sinx\int(\frac{sinx}{cos^{2}x})dx = sinx\int\frac{du}{u^{2}} = sinx(\frac{1}{u}) = \frac{sinx}{cosx} = \tan x$
(By substitute: u = cosx, du= sinxdx)
(c) By (b), we have:
$$y = c_1\sin x + c_2\tan x$$ is the general solution.
then $y' = c_1cosx + c_2sec^{2}x$
Since $y(\frac{\pi}{3})=0, y'(\frac{\pi}{3})=7$
So, $sin(\frac{\pi}{3})c_1 + tan(\frac{\pi}{3})c_2 = 0$ and $cos(\frac{\pi}{3})c_1 + sec^2(\frac{\pi}{3})c_2 = 7$
Then $(\frac{\sqrt{3}}{2})c_1 + \sqrt{3}c_2 = 0$ and $ \frac{c_1}{2} + 4c_2 = 7$
Thus, $c_1= \frac{14}{3}$ and $c_2 = \frac{7}{3}$
Therefore, $y = \frac{14}{3}sinx + \frac{7}{3}tanx$ is a solution to the IVP.

Yulin, please clean you post. Almost impossible to read. \sin, \cos, no \lmoustache but \int
Instead of e^{ some long expression} use \exp\bigl(some long expression}\bigr

I have cleaned my post. Is that OK now? Thanks.

I began to clean, and realized that $W$ calculated incorrectly

I calculated three times again, but I didn't find my mistake. :(

Indeed, it is correct. Simply I did not recognize it