# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Huyi Xiong on October 04, 2019, 02:01:48 PM

Title: TUT 0801 QUIZ2
Post by: Huyi Xiong on October 04, 2019, 02:01:48 PM
\begin{align}
1 + (\frac{x}{y}-sin(y))y' = 0
\end{align}

$M = 1$    $N = \frac{x}{y} - siny$    $M_y = 0$     $N_x = \frac{1}{y}$

$\implies M_y \neq N_x$

\begin{align}
\textbf{Find integrating factor} \\
R_1 &= \frac{M_y-N_x}{M} = \frac{0-\frac{1}{y}}{1} = - \frac{1}{y} \\
\mu &= e^{-\int R_1 dy} = e^{-\int -\frac{1}{y} dy} = e^{lny} = y \\
\textbf{Multiply both sides by $\mu$} \\
y + (\frac{x}{y} - siny)yy' &= 0 \\
y + (x-ysiny)y' &= 0
\end{align}

$P = y$    $Q = x-ysiny$

\begin{align}
f &= \int {P dx} = \int {y dx} = xy + g(y) \\
f_y &= Q = x - ysiny = x + g'(y) \\
g'(y) &= -ysiny \\
g(y) &= -\int {ysiny dy} \\
\textbf{Integrating by parts}
\end{align}

$u = y$    $du = dy$    $v = -cosy$    $dv = sinydy$

\begin{align}
g(y) &= -[-ycosy-\int {-cosy dy]} +c \\
&= ycosy - siny + c \\
\implies f = xy+ycosy-siny=c
\end{align}