MAT244--2018F > Final Exam

FE-P6

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Chengyin Ye:
dx/dt=2x2y+2y3+8y
dy/dt=-2x3-2xy2+32x
so,dx/dy=2x2y+2y3+8y/-2x3-2xy2+32x
so,(2x3+2xy2-32x)dx+(2x2y+2y3+8y)dy=0
Let M=2x3+2xy2-32x,N=2x2y+2y3+8y
My=4xy,Nx=4xy
My=Nx, so Exact.
There exists a 𝐻(𝑥,𝑦) s.t. 𝐻x(𝑥,𝑦)=M,𝐻y(𝑥,𝑦)=N
𝐻(𝑥,𝑦)=1/2x4+x2y2-16x2+h(y)
𝐻y(𝑥,𝑦)=2x2y+ℎ′(𝑦)
so,ℎ′(𝑦)=2y3+8y
h(y)=1/2y4+4y2+C
Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

Tzu-Ching Yen:

--- Quote from: Joyce Ye on December 16, 2018, 07:37:50 PM ---Therefore, 𝐻(𝑥,𝑦)=1/2x4+x2y2-16x21/2y4+4y2=C

--- End quote ---

$+$ missing

Jingze Wang:
Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.

Tzu-Ching Yen:

--- Quote from: Jingze Wang on December 16, 2018, 08:11:04 PM ---Can anyone tell me why Professor said my answer is wrong? Also, to avoid confusion, I use 0.5 instead of 1/2.

--- End quote ---
No idea tbh. Joyce provided exactly the same answer as your corrected version.

Victor Ivrii:
Now it is correct, but before it was not so. Or, may be, because the correct expression was not articulated in the special line, or like this:
$$\boxed{H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).}$$
BTW, making this problem I started from $H(x,y)$ in the form
\begin{align*}
H(x,y)=&\bigl( (x-a)^2 +y^2-b^2\bigr)\bigl( (x+a)^2 +y^2-b^2\bigr)=\\
           &\bigl( x^2 +y^2-b^2+a^2 -2ax\bigr)\bigl( x^2 +y^2-b^2+a^2 +2ax\bigr)=\\
           &\bigl( x^2 +y^2-b^2+a^2\bigr)^2 -4 a^2x^2
\end{align*}
with $b> a>0$.

Question: For $H(x,y)$ what are points $(\pm 4,0)$ and $(0,0)$?

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