Question

i need a demo of the exact DE method thingy... \[2xydx + (y^2 + x^2)dy = 0\]

Answer 1

@Zarkon can you help me here if possible?

Answer 2

integrate one of the "terms"

Answer 3

doesnt it have that partial derivative thingy??

Answer 4

yes

Answer 5

could you teach me how to use that here?

Answer 6

yes, which is your favorite side, right or left?

Answer 7

uhh left

Answer 8

then lets use the left:

integrate 2xy dx

Answer 9

with respect to which variable?

Answer 10

oh x *facepalm*

Answer 11

the dx says to do it to x

Answer 12

yx^2 right?

Answer 13

almost try again

Answer 14

err, you might be right lol

Answer 15

lol =)) so what's next?

Answer 16

2y pulls out as a constant, and x goes to x^2/2 .. yep your good

Answer 17

dont forget your +C tho

Answer 18

i thought C can be put anywhere? o.O i guess this is a different approach...

Answer 19

so it's yx^2 + c

Answer 20

think of the +C as a function of y that gets treated as a constant with respect to x; say + g(y) instead

Answer 21

yx^2 + g(y) ; now take the derivative with respect to y

Answer 22

2xydx?

Answer 23

yx^2 + g(y) wrt y doesnt go back down to "wrt x"

Answer 24

oh sorry

Answer 25

x^2 + 1

Answer 26

oh no...x^2 only

Answer 27

the derivative of g(y) wrt y = g'(y) right?

Answer 28

oh wait...im confused @_@

Answer 29

oh

Answer 30

yeah...makes sense lol...so x^2 + g"(y)?

Answer 31

^that'sone apostrophe only

Answer 32

yx^2 + g(y) wrt y downs to:

x^2 + g'(y) and this should equate to the right side that we left down there

Answer 33

does it?

Answer 34

x^2 + g'(y) = x^2 + y^2

g'(y) = y^2 ; integrate back up to get to g(y)

Answer 35

im confused o.O

Answer 36

why do we want g(y)?

Answer 37

because g(y) is an unknown to us that needs to be determined to find a solution

Answer 38

i see...

Answer 39

so i'll integrate y^2?

Answer 40

yep

Answer 41

y^3 / 3

Answer 42

+C which this time can just be a +C

Answer 43

not g(x) anymore?

Answer 44

we already have a f(x), its what we worked off of to begin with

Answer 45

so...we have g(y) and f(x)...what now?

Answer 46

\[yx^2 + \frac{1}{3}y^3+c\]is our results

Answer 47

the idea is, up one side to find a function that can be down to the other side and compared to

Answer 48

spose we started withe the right side:

integrate x^2 + y^2 dy

Answer 49

y^3/3 + x^2 y

Answer 50

+c, which we regard as +f(x)

Answer 51

derive that wrt x and compare it

Answer 52

2xydx

Answer 53

2xy + f'(x) = 2xy

f'(x) = 0 ; integrate it back up to find f(x)

f(x) = c

Answer 54

either way we go we end up with:\[yx^2 + \frac{1}{3}y^3+c\]

Answer 55

ahh i see..good method

Answer 56

the method our teacher proposed was integrate both and then record all unique terms...seems the same will come out

Answer 57

there are a few methods, im prone to this one tho :)