Author Topic: TT1 Problem 2 (afternoon)  (Read 3022 times)

Victor Ivrii

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TT1 Problem 2 (afternoon)
« 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 ${y(\frac{\pi}{3})=0, y'(\frac{\pi}{3})=7}$.

Yulin WANG

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Re: TT1 Problem 2 (afternoon)
« Reply #1 on: October 16, 2018, 10:23:49 AM »
(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:

$y_2 = y_1\int(\frac{(e^{\int-p(x)dx})}{y_1^{2}})dx$ = $sinx\int(\frac{tan^{2}xsinx}{sin^{2}x})dx$

= $sinx\int(\frac{sinx}{cos^{2}x})dx = -sinx\int\frac{du}{u^{2}} = -sinx(-\frac{1}{u}) = \frac{sinx}{cosx} = \tan x$

(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.
« Last Edit: October 18, 2018, 09:58:55 PM by Victor Ivrii »

Victor Ivrii

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Re: TT1 Problem 2 (afternoon)
« Reply #2 on: October 18, 2018, 04:56:35 AM »
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

Yulin WANG

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Re: TT1 Problem 2 (afternoon)
« Reply #3 on: October 18, 2018, 12:08:52 PM »
I have cleaned my post. Is that OK now? Thanks.

Victor Ivrii

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Re: TT1 Problem 2 (afternoon)
« Reply #4 on: October 18, 2018, 12:50:44 PM »
I began to clean, and realized that $W$ calculated incorrectly

Yulin WANG

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Re: TT1 Problem 2 (afternoon)
« Reply #5 on: October 18, 2018, 05:52:51 PM »
I calculated three times again, but I didn't find my mistake. :(

Victor Ivrii

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Re: TT1 Problem 2 (afternoon)
« Reply #6 on: October 18, 2018, 09:50:16 PM »
Indeed, it is correct. Simply I did not recognize it