Toronto Math Forum

I've modified my answer now 

Describe the locus of points z satisfying the given equation.
\begin{align}
Im(2iz) = 7  \\
z = x + iy,  x,y \in \mathbb{R} \\
Im[2i(x+iy)]&=7 \\
Im(2xi-2y) &= 7 \\
2x &= 7 \\
x &= \frac{7}{2}
\end{align}
The locus of point z is a vertical line cross $(\frac{7}{2}, 0)$.

Solve the initial value problem.
\begin{align*}
y'' -6y' +9y = 0\\
y(0) = 0, y'(0) = 2
\end{align*}

\begin{align}
r^2 - 6r +9 = 0 \\
(r-3)^2 = 0\\
r_1 = r_2 = 3\\\
y(t) = c_1e^{3t}+c_2te^{3t}\\
y(0)=c_1e^0+0=0\\
c_1=0
\end{align}
\begin{align}
y'(0)&=3c_1e^{3t}+c_2(e^{3t}+3te^{3t})\\
&=3c_1e^{3t}+c_2e^{3t}+3c_2 te^{3t} \\
&=3c_1e^0+c_2e^0+0\\
&=3c_1+c_2=2 \\
c_2&=2-3c_1\\
&=2-0=2\\
y(t)&=2te^{3t}
\end{align}

\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}