Question

Match the slope field with the differential equation.

Answer 1

https://i.gyazo.com/27a0d30b69bc4a86b3419438b826eac5.png

Answer 2

@jim_thompson5910 Hook a brotha up with an explaination?

Answer 3

mind if i try :p ?

Answer 4

go ahead @baru

Answer 5

@baru Please do ;)

Answer 6

look at graph C, it is clearly the slope field of a circle
\[x^2+y^2=r^2\]
differentiate that

Answer 7

2x+2y=2r

Answer 8

'r' is a constant so RHS=0

Answer 9

RHS?

Answer 10

right hand side

Answer 11

2x+2y=0

Answer 12

sorry
2x+2y\(\frac{dy}{dx}=0\)

Answer 13

Ohh because we are differentiating by x

Answer 14

So what's next? Separation of variables?

Answer 15

just re-arrange to get
\(\frac{dy}{dx}=?\)
and compare with the options

Answer 16

@Ephemera

Answer 17

It would be -x/y

Answer 18

So that means it would be C for the first one, correct? Due to the circlish look

Answer 19

yep, now we have eliminated graph C and first option

Answer 20

now look at graph B, the slope field cutting the y axis is perfectly horizontal,
or in other words, dy/dx=0 at x=0
which option fits this info?

Answer 21

dy/dx=x/y?

Answer 22

:) yes
it cant be any of the other options, because they have x in the denominator, and division by zero is not defined (the first option would have fit, but we already eliminated that)

Answer 23

followed?

Answer 24

Yeah

Answer 25

I went ahead and attempted the other two

Answer 26

Got
C
D
B
A
in that order

Answer 27

what logic did you use to solve the remaining?

Answer 28

For A I got dy/dx=y/x

Answer 29

Where at (0,0) the slope is positive

Answer 30

:) nice
that looks right

Answer 31

Thanks for the help.

Answer 32

woops, you have got the right answer, but \(\frac{0}{0}\) is not defined
so we cannot eliminate based on that

Answer 33

this is what I got

Answer 34

@jim_thompson5910 i agree

@Ephemera
graph D looks like the upward hyperbola
\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\] where and b are constants

Answer 35

*a and b

Answer 36

use the same procedure as for the circle and find dy/dx

Answer 37

Oh ok....I should've chosen a different point to test as well should've known it was undetermined which would mislead to wrong answer. Thanks for the help.

Answer 38

@jim_thompson5910 you have a better method? rather than guessing the graph?

Answer 39

why not solve for y and see what forms you get

for example...

\[\Large \frac{dy}{dx} = -\frac{x}{y}\]

\[\Large y dy = -x dx\]

\[\Large \int y dy = \int -x dx\]

\[\Large \frac{y^2}{2} = -\frac{x^2}{2}+C\]

\[\Large y^2 = -x^2+2C\]

\[\Large y^2 = -x^2+C\]

\[\Large x^2 + y^2 = C\]

Equation for a circle.

So \(\Large \frac{dy}{dx} = -\frac{x}{y}\) matches with C. The slope field is basically a group of concentric circles

Answer 40

slopefield D is actually a bunch of lines (not hyperbolas)

Answer 41

B is the hyperbolic one

Answer 42

hmm...yep, y'=y/x is diff eq for the special case hyperbola of a=b,not any hyperbola, my mistake :P

Answer 43

\(y^2-x^2=a^2\\2yy'-2x=0\\y'=x/y\)
hmmm, i'm confused, straight lines and the spcl case hyperbola fit, any clue in the graph to suggest one over the other?

Answer 44

nevermind xD i see my error