Toronto Math Forum

$M = 2x \sin(y) +1 $

$N = 4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y) $

$M_y = 2x\cos(y)$

$N_x = 8x\cos(y) + 3cos(y)$

$\frac{M_y - N_x} {M} = \frac{-6x\cos(y) - 3cos(y)}{2x\sin(y) + 1} = -3\cot(y)$

$\mu = e^{-\int -3\cot(y) dy } = \sin^3(y)$

Then multiply $sin^3(y)$ on M and N.

Then we get $M =  sin^3(y)(2x \sin(y) +1)$, $N =sin^3(y)(4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y))$

$M_y = \sin^2(y) \cos(y) (8x \sin(y) + 3) $, and $N_x = \sin^2(y) \cos(y) (8x \sin(y) + 3) $

Thus $M_y = N_x$, exact.

Then $\psi_x = M$

$\psi = \int M dx = \int \sin^3(y)(2x \sin(y) +1) dx = \sin^3(y)(x^2\sin(y) + x) + h(y) $

$N = \psi_y = 4x^2\sin^3(y)\cos(y) + x\sin^2(y)\cos(y) + h'(y) $

$h'(y) = 10\sin^4cos(y) $

$h(y) = \int 10\sin^4cos(y) dy = 2\sin^5(y) + c$

Then $\psi = \sin^3(y)(x^2\sin(y) + x) + 2\sin^5(y) = c$

$$\tan(z) = \frac{\sin(z)}{\cos (z)} = \frac{z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...}{1 - \frac{z^2}{2!}+\frac{z^4}{4!}...} = a_0 + a_1z + a_2z^2 +a_3z^3 + ...$$

Thus
$(a_0 + a_1z + a_2z^2 +a_3z^3 + ...)(1 - \frac{z^2}{2!}+\frac{z^4}{4!}...) = z - \frac{1}{3!}z^3 + \frac{z^5}{5!} - \frac{z^7}{7!}...$

Thus we get

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

.
.
.

Therefore $\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 ....$

\begin{align*}
 \frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
 &= 1- \frac{1}{z+1+2}\\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
 &= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
 &= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
 &= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}

homogeneous equation is
$y^{'''} - y^{'} = 0$

then $r^3 - r = 0$

$r = 0, \pm 1$

thus, $y_{c}(t) = c_{1} + c_{2}e^t + c_{3}e^{-t}$

\begin{align*}
W(t) &=

\begin{bmatrix}
1 & e^t & e^{-t} \\
0 & e^t & -e^{-t}\\
0 & e^t & e^{-t}
\end{bmatrix} \\
\\

&= 1*(-1)^{1+1}
\begin{bmatrix}
e^t & -e^{-t}\\
e^t & e^{-t}
\end{bmatrix}
+ e^t(-1)^{2+1}
\begin{bmatrix}
0 & -e^{-t}\\
0 & e^{-t}
\end{bmatrix}
+ e^{-t}(-1)^{3+1}
\begin{bmatrix}
0 & e^t\\
0 & e^t
\end{bmatrix}\\
\\

&= e^t*e^t - (-e^{-t})e^t + 0 + 0 \\
\\

&= 1 + 1 \\
\\

&= 2
\end{align*}

\begin{align*}
W_{1}(t) &=
\begin{bmatrix}
0 & e^t & e^{-t}\\
0 & e^t & -e^{-t}\\
1 & e^t & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+3}
\begin{bmatrix}
e^t & e^{-t}\\
e^t & -e^{-t}
\end{bmatrix}\\
\\

&= -1-1 \\
\\

&= -2
\end{align*}

\begin{align*}
W_{2}(t) &=
\begin{bmatrix}
1 & 0 & e^{-t}\\
0 & 0 & -e^{-t}\\
0 & 1 & e^{-t}
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
0 & -e^{-t}\\
1 & e^{-t}
\end{bmatrix}\\
\\

&= e^{-t}
\end{align*}
\\

\begin{align*}
W_{3}(t) &=
\begin{bmatrix}
1 & e^t & 0\\
0 & e^t & 0\\
0 & e^t & 1
\end{bmatrix}\\
\\

&=1*(-1)^{1+1}
\begin{bmatrix}
e^{t} & 0\\
e^{t} & 1
\end{bmatrix}\\
\\

&= e^{t}
\end{align*}

\begin{align*}
y_{p}(t) &= y_{1}\int \frac{g(t)W_{1}(t)}{W(t)} dt + y_{2}\int \frac{g(t)W_{2}(t)}{W(t)} dt + y_{3}\int \frac{g(t)W_{3}(t)}{W(t)} dt \\
\\

&= 1\int \frac{\cosh(t)(-2)}{2)} dt + e^t\int \frac{\cosh(t)e^{-t}}{2} dt + e^{-t}\int \frac{\cosh(t)e^t}{2} dt \\
\\

&= -1\sinh(t) + \frac{1}{2}e^t\int \cosh(t)e^{-t} dt + \frac{1}{2}e^{-t}\int \cosh(t) e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t \int \cosh(t)e^{-t}dt + \frac{1}{2}e^{-t}\int \cosh(t)e^t dt \\
\\

&= -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})
\end{align*}

Therefore $y(t) = c_{1} + c_{2}e^t + c_{3}e^{-t} -\sinh(t) + \frac{1}{2}e^t(\frac{t}{2} - \frac{e^{-2t}}{4}) + \frac{1}{2}e^{-t}(\frac{t}{2}+ \frac{e^{2t}}{4})$

sorry, there is a typo. The solution should be the given function doesn't represent a locally sourceless and an irrotational flow 

$$g(z) = \frac{4z^6-7z^3}{(Z^2-4)^3} = \frac{4z^6-7z^3}{z^6-12z^4+48z^2-64}$$
we can divide $z^6$ on both numerator and denominator.
Then we can get
$$g(z) = \frac{4-7z^{-1}}{z-12z^{-2}+48z^{-4}-64z^{-6}}$$
Then as $z\to\infty $
$$\lim_{z\to\infty} f(z) = \frac{4 - 0}{1 - 0 + 0 - 0} = 4$$