Author Topic: Problem 1 (afternoon)  (Read 844 times)

Victor Ivrii

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Problem 1 (afternoon)
« on: October 23, 2019, 05:57:42 AM »
(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
-y^2\sin(xy) + \bigl(-xy \sin(xy)+2\cos(xy)+3y\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(\dfrac{\pi}{3})=1$.

Hongling Liu

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Re: Problem 1 (afternoon)
« Reply #1 on: October 23, 2019, 06:42:50 AM »
-y^2•sinxy + (-xy•sinxy + 2cosxy + 3y)•y’ = 0
Solution:
a):
My = -2y•sinxy - x•y^2•cosxy
Nx = -3y•sinxy - x•y^2•cosxy
My≠Nx
∴R1 = (My - Nx)/M = -1/y
u = e^(∫-R1dy) = y
∴ -y^3•sinxy + y•(-xy•sinxy + 2cosxy + 3y)•y’ = 0
My = -3y^2•sinxy - x•y^3•cosxy
Nx = -3y^2•sinxy - x•y^3•cosxy
∴My =Nx
ψ(x,y) = ∫Mdx = y^2•cosxy + h(y)
ψy = N
ψy = 2y•cosxy - x•y^2sinxy + h’(y)
∴h’(y) = 3y^2
  h(y) = y^3
ψ(x,y) = y^2•cosxy + y^3 = C
b):
y(π/3) = 1 ∴C= 3/2
∴ ψ(x,y) = y^2•cosxy + y^3 = 3/2

Correct but looks really ugly . V.I.
« Last Edit: October 31, 2019, 09:35:05 AM by Victor Ivrii »

Lan Cheng

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Re: Problem 1 (afternoon)
« Reply #2 on: October 23, 2019, 06:53:28 AM »
a) Let $M=-y^{2}sin(xy),N=-xysin(xy)+2cos(xy)+3y,$

$My=-2ysin(xy)-xy^{2}cos(xy), Nx=-ysin(xy)-xy^{2}cos(xy)-2ysin(xy).$ Should be $M_y$ and so on

Let $R_{1}=\frac{My-Nx}{M}=\frac{ysin(xy)}{-y^{2}sin(xy)}=-\frac{1}{y},$

$\mu=e^{-\int R_{1}dy}=e^{\int\frac{1}{y}dy}=e^{ln(y)}=y.$

multiply each side by y,

$-y^{3}sin(xy)+(-xy^{2}sin(xy)+2ycos(xy)+3y^{2})y'=0.$

$My=-3y^{2}sin(xy)-xy^{3}cos(xy),Nx=-3y^{2}sin(xy)-xy^{3}cos(xy)=My.$

Thus, there exist $\varphi(x,y) such that \varphi x=M,\varphi y=N.$

$\varphi=y^{2}cos(xy)+h(y),\varphi y=2ycos(xy)-xy^{2}sin(xy)+h'(y)=-xy^{2}sin(xy)+2ycos(xy)+3y^{2}.$

$h'(y)=3y^{2},h(y)=y^{3}+C.$

Thus, $y^{2}cos(xy)+y^{3}=C.$

b) $y(\frac{\pi}{3})=1,cos(\frac{\pi}{3})+1=C.$

$C=\frac{3}{2},y^{2}cos(xy)+y^{3}=\frac{3}{2}.$
« Last Edit: October 31, 2019, 09:39:09 AM by Victor Ivrii »

huoyanro

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Re: Problem 1 (afternoon)
« Reply #3 on: October 23, 2019, 06:57:57 AM »
(a):
My = -2y•sin(xy) - x•y^(2)•cos(xy)
Nx = -3y•sin(xy) - x•y^(2)•cos(xy)
My≠Nx
not exact
R1 = (Nx-My)/M = 1/y
u = e^(∫R1dy) = y
-y^3•sin(xy) + y•(-xy•sin(xy) + 2cos(xy) + 3y)•y’ = 0
My = -3y^2•sin(xy) - x•y^3•cos(xy)
Nx = -3y^2•sin(xy) - x•y^3•cos(xy)
My =Nx
exact
there exists ψ(x,y) such that ψx= ∫Mdx = y^2•cos(xy) + h(y)
ψy = N
ψy = 2y•cos(xy) - x•y^2sin(xy)+ h’(y)
h’(y) = 3y^2
h(y) = y^3
ψ(x,y) = y^2•cos(xy) + y^3 = C
b):
y(π/3) = 1 ∴C= 3/2
ψ(x,y) = y^2•cos(xy) + y^3 = 3/2

yangyiq5

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Re: Problem 1 (afternoon)
« Reply #4 on: October 23, 2019, 07:30:09 AM »
$$M_{y} = -2ycos(xy) + xy^{2}sin(xy)$$
$$N_{x} = -ycos(xy) + xy^{2}sin(xy) - 2ycos(xy)$$
$$R_{2} = \frac{M_{y}-N_{x}}{M} = \frac{-2ycos(xy) + xy^{2}sin(xy)-(ycos(xy) + xy^{2}sin(xy) -2ycos(xy))}{-y^{2}cos(xy)}$$
$$R_{2} = -\frac{1}{y}$$
$$\mu = e^-{\int -\frac{1}{y}} = y$$
$$multiple at both sides$$
$$-y^{3}cos(xy)+(-xy^{2}cos(xy)-2ysin(xy)+3y^{2})y’ = 0$$
$$\exists \varphi$$
$$\varphi _{x} = M $$
$$\varphi _{y} =N$$
$$\varphi  = \int -y^{3}cos(xy) dx = -y^{2}sin(xy) + h(y)$$
$$\varphi_{y} = -2ysin(xy)-xy^{2}cos(xy) + h’(y)= -xy^{2}cos(xy)- 2ysin(xy)+3y^{2}$$
$$h’(y) = 3y^{2}$$
$$h(y) = y^{3} + C$$
$$\varphi = -y^{2}sin(xy) + y^{3} + C$$
$$when x = n/3  y = 1  $$
$$C = 3/2$$

Yingyingz

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Re: Problem 1 (afternoon)
« Reply #5 on: October 23, 2019, 07:45:44 AM »
(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

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Re: Problem 1 (afternoon)
« Reply #6 on: October 23, 2019, 11:52:18 AM »
\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 $

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

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