Question

HELP!

Answer 1

i got C @asnaseer

Answer 2

@AMYCARTER

Answer 3

@SithsAndGiggles PLEASE HELP

Answer 4

i just need someone to check my answer please

Answer 5

Have you tried solving any of these ODEs? They are all separable, so not too difficult.

Answer 6

Before you try that, though, I'm sure there's an easier (but somewhat tedious) way to approach this. Pick some points from the graph, like the \((1,1)\) and \((-1,-1)\), for example, and plug them into the DE. This way you can eliminate which ODEs are NOT satisfied.

For example, take \((-1,-1)\). You get
\[\begin{matrix}\frac{dy}{dx}_A=\frac{-1}{(-1)^2}=-1&&\frac{dy}{dx}_B=\frac{-1}{-1}=1\\\\
\frac{dy}{dx}_C=\frac{(-1)^2}{-1}=-1&&\frac{dy}{dx}_D=\frac{(-1)^2}{(-1)^2}=1
\end{matrix}\]
According to the slope field, at \((-1,-1)\), the slope appears to be negative, so you can eliminate B and D.

Answer 7

thats such a more simple way to do it...i took up like a whole page

Answer 8

but i think we both got the same answer whih was C

Answer 9

Yes, that's right. If you try the point \((-1,1)\) or \((1,-1)\), you'll eliminate A.

Answer 10

thanks so much(:

Answer 11

it would be parabolas right?

Answer 12

@SithsAndGiggles

Answer 13

Not parabolas, no. You can tell by solving - there are a few ways, but treating this as a separable is easiest.
\[\begin{align*}x\,dx+y\,dy&=0\\
y\,dy&=-x\,dx\\
\int y\,dy&=-\int x\,dx\\
\frac{1}{2}y^2&=-\frac{1}{2}x^2+C\\
y^2+x^2&=C^*
\end{align*}\]
Which conic is this?

Answer 14

hyperbola @SithsAndGiggles

Answer 15

No, not hyperbola

Answer 16

oh wait circle

Answer 17

right? @SithsAndGiggles

Answer 18

Yes, a circle. Remember, \(x^2+y^2=r^2\) is a circle with radius \(r\).

Answer 19

thanks for the help(: