### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Yufang Liu

Pages: [1]
1
##### Term Test 1 / Re: TT1 Problem 1 (main)
« on: October 16, 2018, 08:27:01 AM »
Here is my solution
since $$M_y=(x^2cos(y) - sin(y))_y = -x^2sin(y) - cos(y)$$
and $$N_x = (xcos(y) - x^3 sin(y))_x = cos(y) - 3x^2siny \neq M_y$$
Notice $$\frac{M_y - N_x}{N} = \frac{2x^2siny - 2cos(y)}{xcos(y) - x^3 sin(y)} =-\frac{2}{x} = f(x)$$
Let the integration factor be $u = u(x)$
$$(Mu)_y = uM_y$$
$$(Nu)_x = u'N + uN_x$$
$$(Mu)_y = (Nu)_x \Longrightarrow \frac{u'}{u} = \frac{M_y - N_x}{N} = f(x)$$
so $$ln u = -\int\frac{2}{x} = ln(x^2) + C$$
so the integration factor is $u(x) = x^{-2}$
Therefore $$F(x, y) = \int(Mu)dx = xcos(y) + \frac{siny}{x} + g(y) + C$$
$$F(x, y) = \int(Nu)dy = xcos(y) + \frac{siny}{x} + C$$
So the general solution is $F(x, y) = 0$ i.e. $$xcos(y) + \frac{siny}{x} + C = 0$$
plug in $y(1) = \pi \Longrightarrow C = 1$
so the solution is $$xcos(y) + \frac{siny}{x} + 1 = 0$$

Pages: [1]