MHB Get the right answer but intermediate steps may be wrong

[find_the_fun]

I am able to solve problems of the kind, but I don't understand what I'm doing :D

Problem: determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point [math](x_o, y_o)[/math] in the region.

[math](4-y^2)y'=x^2[/math]

My work:
[math]\frac{\partial f}{\partial y}=\frac{2x}{4-y^2}[/math]
$$\therefore y\ne\pm2$$

[math]\frac{\partial f}{\partial x}=\frac{4}{4-y^2}[/math]
$$\therefore y\ne\pm2$$

Answer key:
the reigions defined by y>2, y < -2 or -2<y<2
*other questions of the type of answers such asa as "halfplanes defined by either y>0 or y<0" this I don't understand; what's a halfplane?

Is it correct how I write [math]\frac{\partial f}{\partial y}[/math] and [math]\frac{\partial f}{\partial x}[/math], I'm not sure if it should be f or something else.

The question asks to find a region. When stating a region is interval notation always used?

 

[Deveno]

I am able to solve problems of the kind, but I don't understand what I'm doing :D

Problem: determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point [math](x_o, y_o)[/math] in the region.

[math](4-y^2)y'=x^2[/math]

My work:
[math]\frac{\partial f}{\partial y}=\frac{2x}{4-y^2}[/math]
$$\therefore y\ne\pm2$$

[math]\frac{\partial f}{\partial x}=\frac{4}{4-y^2}[/math]
$$\therefore y\ne\pm2$$

Answer key:
the reigions defined by y>2, y < -2 or -2<y<2
*other questions of the type of answers such asa as "halfplanes defined by either y>0 or y<0" this I don't understand; what's a halfplane?

Is it correct how I write [math]\frac{\partial f}{\partial y}[/math] and [math]\frac{\partial f}{\partial x}[/math], I'm not sure if it should be f or something else.

The question asks to find a region. When stating a region is interval notation always used?

I have no idea what "$f$" is.

We have:

$y' = \dfrac{x^2}{4 - y^2}$

This doesn't tell us what function of $x$ that $y$ is, but it DOES tell us that $y$ cannot be $\pm 2$ (or else $y'$ is undefined).

A half-plane is exactly what it sounds like: half of the plane; that is, two neighboring quadrants. There are four such beasts:

$\{(x,y) \in \Bbb R^2: x > 0\}$ (the "right half-plane")
$\{(x,y) \in \Bbb R^2: x < 0\}$ (the "left half-plane")
$\{(x,y) \in \Bbb R^2: y > 0\}$ (the "upper half-plane")
$\{(x,y) \in \Bbb R^2: y < 0\}$ (the "lower half-plane")

A region is a subset $U$ of the plane with the following properties:

1) Some point $(x,y)$ lies in the region $U$.
2) $U$ is a connected set.
3) $U$ is OPEN.

Some authors allow boundary points of $U$ to be added to $U$, and also call this a region. Examples of regions:

$\{(x,y) \in \Bbb R^2: x^2 + y^2 < 1\}$ -the open unit disk

$(a,b) \times (c,d)$ -an open RECTANGLE with corners $(a,c),(b,c),(a,d),(b,d)$ (note: the first set of pairs- in $(a,b)\times(c,d)$ refer to open intervals, the second set of pairs refer to POINTS).

$(-a,a) \times \Bbb R$-an open STRIP of width $2a$ centered on the $y$-axis

$\{(x,y) \in \Bbb R^2: xy < 1\}$ -the region bounded by the hyperbola $xy = 1$. Note this region contains all of the second and fourth quadrants.

The practical value of regions is this: for any point INSIDE a region, we can find a suitably small disk (or rectangle, depending on one's approach) entirely in the region. This let's us take limits, such as derivatives, and check for continuity.

So, no, regions do not HAVE to be defined in terms of intervals-but...we can use "interval arguments" as long as we are "close enough" to a point $(x_0,y_0)$ in the region (boundary points usually require special consideration).