MAT244--2019F > Term Test 1

Problem 1 (afternoon)

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Yingyingz:
(a)$$
\begin{array}{l}{M=-y^{2} \sin (x y)} \\ {M y=-\left(2 y \sin (x y)+y^{2} \cos (x y) \cdot x\right)} \\ {N=-x y \sin (x y)+2 \cos (x y)+3 y} \\ {N x=-(y \sin (x y)+x y \cos (x y) \cdot y)-2 \sin (x y) y}\end{array}
$$
$$
M y \neq N_x
$$
$$
R_{1}=\frac{M y-N x}{M}=\frac{-y \sin (x y)+2 \sin (x y) y}{-y^{2} \sin (x y)}=-\frac{1}{y}
$$
$$
\therefore \mu=e^{-\int-R_{1} d y}=e^{\int \frac{1}{y} d y}=y
$$

Multiply $\mu$ on both sides,
$$
-y^{3} \sin (x y)+\left(-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}\right) y^{\prime}=0
$$
$$
\begin{array}{l}{M^{\prime}=-y^{3} \sin (x y)} \\ {M^{\prime} y=-\left(3 y^{2} \sin (x y)+y^{3} \cos (x y) x\right)} \\ {N^{\prime}=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}} \\ {N^{\prime} x=-\left(y^{2} \sin (x y)+x y^{2} \cos (x y) y\right)-2 y \sin (x y) y} \\ {\quad=-3 y^{2} \sin (x y)-x y^{3} \cos (x y)}\end{array}
$$
$$
\therefore M^{\prime} y=N^{\prime} x
$$
$\therefore \exists \varphi(x, y), s, t$
$$
\begin{array}{l}{\varphi_{x}=M^{\prime}=-y^{3} \sin (x y)} \\ {\varphi=\int M^{\prime} d x=y^{2} \cos (x y)+h(y)} \\ {\varphi_{y}=2 y \cos (x y)-y^{2} \sin (x y) x+h^{\prime}(y)}\end{array}
$$
$$
\because N^{\prime}=\varphi y=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}
$$
$$
\begin{array}{l}{\therefore h^{\prime}(y)=3 y^{2}} \\ {h(y)=\int h^{\prime}(y) d y=y^{3}+C}\end{array}
$$
$$
\therefore \varphi=y^{2} \cos (x y)+y^{3}=c
$$
(b) $$y\left(\frac{\pi}{3}\right)=1$$
$$
\begin{aligned} \therefore \quad x &=\frac{\pi}{3}, \quad y=1 \\ c &=y^{2} \cos (x y)+y^{3} \\ &=\frac{1}{2}+1 \\ &=\frac{3}{2} \end{aligned}
$$
$$
\therefore \varphi=y^{2} \cos (x y)+y^{3}=\frac{3}{2}
$$

Ranran Wang:
\textbf{\textbf{Ans:}}

$M=-y^{2} \sin (x y)$

$N=-x y \sin (x y)+2 \cos (x y)+3 y$

$M_{y}=-2 y \sin (x y)-y^{2} \cos (x y) \cdot x=-2 y \sin (x y)-x y^{2} \cos (x y)$

$N_{x}=-y \sin (x y)-x y^{2} \cos (x y)+2(-\sin (x y)) y+0=-y \sin (x y)-x y^{2}(\cos (x y)-2 y \sin (x y)$

$R_1=\frac{M_{y}-N_{x}}{M}=\frac{-2 y \sin (x y)-x y^{2} \cos (x y)+y \sin (x y)+x y^{2} \cos (x y)+2 y \sin (x y)}{-y^{2} \sin (x y)}=\frac{y \sin (x y)}{-y^{2} \sin (x y)}=-\frac{1}{y}$

$\mu=e^{-\int_{R_1} d y}=e^{\int \frac{1}{y} d y}=e^{\ln |y|}=y$

both muliply by $\mu$ $-y^{3} \sin (x y)+\left(-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}\right) y^{\prime}=0$

$\varphi(x, y)$

$\varphi_{x}=M=-y^{3} \sin (x y)$

$\varphi_{y}=N=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}$

$\varphi=\int M d x=\int-y^{3} \sin (x y) d x=y^{2} \cos (x y)+h(y)$

$\varphi_{y}=2 y \cos (x y)+y^{2} x \sin (x y)+h^{\prime}(y)=N$

$h^{\prime}(y)=3 y^{2}$

$\int h^{\prime}(y) d y=\int 3 y^{2} d y=y^{3}$

$\varphi=y^{2} \cos (x y)+y^{3}=C$

put $y\left(\frac{\pi}{3}\right)=1$

$x=\frac{\pi}{3}$

$y=1$  into $ \varnothing $

$1^{2} \cos \left(\frac{\pi}{3} \times 1\right)+1^{3}=C=\frac{1}{2}+1=\frac{3}{2}$

$\therefore y^{2} \cos (x y)+y^{3}=\frac{3}{2}$

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