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- anonymous

The general solution of the differential equation dy − 0.2x dx = 0 is a family of curves. These curves are all
lines
hyperbolas
parabolas
ellipses

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- anonymous

- chestercat

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- anonymous

http://www.wolframalpha.com/widgets/view.jsp?id=e602dcdecb1843943960b5197efd3f2a

- anonymous

i'm thinking hyperbolas but not sure

- anonymous

\[dy=0.2x dx\]
by moving over
\[\int\limits dy=\int\limits 0.2x dx\]
Is that what you did?

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- anonymous

and then I got y=0.2 which is a graph of a straight line at y=0.2, so I would call it a line.

- anonymous

um .. no i just shamelessly plugged it into wolfram :P

- anonymous

Does this involve the slope field?

- anonymous

it might, some of the other questions iv been asked have involved the slope field

- anonymous

plug the equation into wolfram and scroll down to the family curve , it seems to show a parabola or hyperbola

- Michele_Laino

after a simple integration I got this:
\[\Large \int {dy = 0.2\int {xdx} = 0.2 \cdot \frac{{{x^2}}}{2}} + C\]

- anonymous

Seems like a parabola. Any 2 curves symmetrical is a hyperbola.

- anonymous

oh yea. I took derivative instead of integrating it..

- anonymous

okay can you just explain to me what the hell family curves are ?

- Michele_Laino

since C is an arbitrary real constant, whose values can vary inside the set of real numbers, in other words, we have:
\[\Large C \in \mathbb{R}\]

- Michele_Laino

|dw:1439275918663:dw|

- anonymous

Family curves are a set of curves.
If the function C is a constant value added to all the family curves, that means all of the curves have the same behavior right?

- anonymous

oh okay so basically its what the graph would look like despite what c could be, right?

- Michele_Laino

yes! right! they differ only for the value of C

- anonymous

If the function y C*
I mean.

- anonymous

okay and this graph would look like a parabola despite what c is right?

- anonymous

Since the graph shows the curves not symmetrical, therefore it cannot be a parabola.

- anonymous

I mean it cannot be a hyperbola.

- Michele_Laino

right! @Jdosio

- anonymous

okay thank you guys very much :D

- Michele_Laino

:)

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