The slope field for a differential equation is shown in the figure. Determine the general solution of this equation

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

answer choices
y=Cx^2
x=Cy^2
x^2 – y^2 = C^2
x2 + y2 = C^2

- anonymous

okay so i know i have to reverse engineer this but i'm not doing too well

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## More answers

- ganeshie8

|dw:1439493039926:dw|

- anonymous

so i should just isolate the variables of each of the answer choices and derive each side right and graph all the diff equations i get and match it to the one im given , is that right?

- Astrophysics

Could you not let x = 1 and y = 1 and then set it up as dy/dx that way you can see the slope easier?

- ganeshie8

To my knowledge that slope field corresponds to a differential equation :
\[y'=2x\]
the family of solutions must look something like
\[y = x^2+C\]

- anonymous

http://www.mathscoop.com/calculus/differential-equations/slope-field-generator.php
this might help

- anonymous

and yes you are correct @ganeshie8 i just cant get the differential equation when onlly given the solution to a differential equation

- anonymous

okay can you walk me through your answer ?

- Astrophysics

Wait, I don't know if it's right, but what I was thinking is as I said above, let x = 1, y = 1 or what ever and just let dy/dx = x/y^2 and then plug in the values and just observe the points...not sure if that's a good method haha, ganeshie definitely knows more.

- ganeshie8

the solution curves are the smooth curves obtained by joining the slope segments in the slope field :
|dw:1439494028215:dw|

- ganeshie8

Easy to see that they are family of parabolas that open up, under vertical translation

- anonymous

okay let me rephrase the question, if i give you the solution to a differential equation (say x=Cy^2) can you find the differential equation it came from ?

- anonymous

lol i know i dont know how to make the constant multiply :P

- ganeshie8

x = Cy^2
1 = C2yy'
x/1 = y/(2y')
y' = y/(2x)

- anonymous

okay and if we graph that we get

##### 1 Attachment

- anonymous

so b is not our answer, so know tell me how you did that so i can do it to the rest of the answer choices XD

- ganeshie8

since you have "one" arbitrary constant, you want to differentiate the solution just once

- anonymous

okay
so from x = Cy^2
to 1 = C2yy'
the only thing you did was derive each side right ?

- ganeshie8

right, just implicitly differentiate both sides with respect to x

- anonymous

so what did you do from 1 = C2yy'
to x/1 = y/(2y')
?

- ganeshie8

x = Cy^2 ---------------(1)
1 = C2yy' --------------(2)
next divide : (1)/(2)

- anonymous

okay umm why ?

- ganeshie8

your goal is to get rid of that arbitrary constant C
dividing both equations does just that

- Astrophysics

Nice explanation @ganeshie8

- anonymous

okay i see so what do you do after you vet rid of the c ?

- anonymous

get*

- ganeshie8

you're done! the equation w/o arbitrary constant is the mother "differential equation" that represents the "family of solution curves" : x = Cy^2

- anonymous

okay so let me try to do that for a different answer choice and tell me if i'm way off.

- ganeshie8

the differential equation y' = y/(2x) represents the whole family of curves x = Cy^2

- ganeshie8

Btw, none of the options are correct, but if I were you, I would tick option A

- anonymous

um okay what makes you chose option 1 ?

- ganeshie8

The correct answer is \(y = x^2 + C\)
From the slope field it is clear that we must get solution curves that look same upto a vertical translation. My guess is that your professor cooked up this problem carelessly... Option 1 is close to parabolas, so..

- anonymous

okay sounds about right :P , will let you know if its right in a little while when i turn it in

- ganeshie8

good luck! please do share what the grader thinks is right

- anonymous

will do ! thanks for all your help guys ! :D

- IrishBoy123

you can re-write these things as level surfaces and i think you can nail it pretty solidly
so if you take option B that has been discounted, you can say: \(x=Cy^2 \implies f(x,y) = x - Cy^2 = constant\)
\(grad f = <1, -2Cy>\)
so the slope \(dy/dx = \frac{-1}{-2Cy} = \frac{1}{2Cy}\)
and it can't be that because dy/dx has no dependency on y and doesn't go ballistic at y = 0

- anonymous

okay are you suggesting its option 1 then ?

- IrishBoy123

i'd go A
i can justify it, just lost a load of latex, so i might try post it again later

- IrishBoy123

here's the observation:
if you write \(y = mx + c\) as a closed surface , you get \(f(x,y) = y - mx = const\)
so \(grad(f) = <-m, 1>\)
so slope in normal xy-speak is \(-\frac{f_x}{f_y}\)
that's not that surprising as \(df = f_x dx + f_y dy = 0 \implies \frac{dy}{dx} = -\frac{f_x}{f_y}\)
do that for \(y=Cx^2\) and i think you get something that behaves like your graph.

- anonymous

i was right !! :D thanks again !!

- IrishBoy123

i reckon the "actual" right answer is:
\(y=\frac{x^2}{2}+C \)

- anonymous

i mean it not i

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