# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:22:58 AM

Title: TT1 Problem 1 (main)
Post by: Victor Ivrii on October 16, 2018, 05:22:58 AM
Find integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(x^2\cos(y)-\sin(y)\bigr) + \bigl(x\cos(y)-x^3\sin(y)\bigr) y'=0.
\end{equation*}

Also, find a solution satisfying $y(1)=\pi$.
Title: Re: TT1 Problem 1 (main)
Post by: Yufang Liu 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 $$xcos(y) + \frac{siny}{x} + 1 = 0$$
Title: Re: TT1 Problem 1 (main)
Post by: Shuhan Lin on October 17, 2018, 08:39:03 PM
here is my hand-written answer for question 1, hope it works
Title: Re: TT1 Problem 1 (main)
Post by: Victor Ivrii on October 18, 2018, 04:22:09 AM
Yufang did everything right -- typing is not perfect -- you need to escape sin, cos etc: \sint t , resulting in  $\sin t$ (upright and a proper spacing).

Shuhan what was the reason to post? Especially in colour?