Question

Find equation of curve which passes through (3,-4) and has slope 2y/x at the point (x,y) on it. I have the answer, just need to learn the logic...

Answer 1

y+4=2y/x(x-3)
y = 2y-6y/x -4
-y=-6y/x -4
y = 6y/x+4

Answer 2

different from answer in book. for context, this is in a chapter called "Applications of first-order differential equations"

Answer 3

the answer is 9y + 4x^2 = 0

Answer 4

Anybody know how to get this answer?

Answer 5

Looks a little peculiar, because if the curve is 9y + 4x^2 = 0, then dy/dx = -8x/9.

Answer 6

Hmm, but -8x/9 does equal 2y/x, so I guess that's legit.

Answer 7

Ok. yeah. Any advice on how to get to that answer?

Answer 8

The way I would normally do it is to separate and integrate.
If dy/dx = 2y/x, then 2dx/x = dy/y.

Answer 9

that's the direction I went too. That got me to y = x^2 + c

Answer 10

Ah, you're missing a constant multiple on y.

Answer 11

Integrating 2dx/x = dy/y -> ln(x^2)=ln(y)+c
exponentiate both sides ->x^2 = cy

Answer 12

Can you get it from here @motza11 ?

Answer 13

calculating...

Answer 14

so integrating dy/y = 2/x dx, I get y = cx^2

Answer 15

which is the same as yours since I can make "c" anything I want.. including 1/c...

Answer 16

Yes.

Answer 17

now what, do I plug in (3,-4) to get the correct c?

Answer 18

That's it.
Then the last steps are to rearrange things to make it look pretty, and then to sit back and admire your work.

Answer 19

BOOM!!!

Answer 20

Thanks, Cliff!

Answer 21

You're welcome. ;-)