MAT244--2019F > Term Test 1

Problem 1 (afternoon)

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Victor Ivrii:
(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:
-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.

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

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

huoyanro:
(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:
$$M_{y} = -2ycos(xy) + xy^{2}sin(xy)$$
$$N_{x} = -ycos(xy) + xy^{2}sin(xy) - 2ycos(xy)$$
$$R_{2} = \frac{M_{y}-N_{x}}{M} = \frac{-2ycos(xy) + xy^{2}sin(xy)-(ycos(xy) + xy^{2}sin(xy) -2ycos(xy))}{-y^{2}cos(xy)}$$
$$R_{2} = -\frac{1}{y}$$
$$\mu = e^-{\int -\frac{1}{y}} = y$$
$$multiple at both sides$$
$$-y^{3}cos(xy)+(-xy^{2}cos(xy)-2ysin(xy)+3y^{2})y’ = 0$$
$$\exists \varphi$$
$$\varphi _{x} = M $$
$$\varphi _{y} =N$$
$$\varphi  = \int -y^{3}cos(xy) dx = -y^{2}sin(xy) + h(y)$$
$$\varphi_{y} = -2ysin(xy)-xy^{2}cos(xy) + h’(y)= -xy^{2}cos(xy)- 2ysin(xy)+3y^{2}$$
$$h’(y) = 3y^{2}$$
$$h(y) = y^{3} + C$$
$$\varphi = -y^{2}sin(xy) + y^{3} + C$$
$$when x = n/3  y = 1  $$
$$C = 3/2$$

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