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« 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.