Question

how to use separable equations to solve:
y'(x)=(-2y(x)+19 )^0.5 and y(-1)=-3
then y(x)=?

Answer 1

take y(x) = (-2y(x) + 19)^(1/2)

so we have

y(-1) = (-2*y(-1) + 19)^(1/2)
and we know y(-1) = -3

can you get it from here?

Answer 2

no sry its y'(x)

Answer 3

hmm, something odd...
you are saying sqrt of something = -3

Answer 4

oh i c, one sec

Answer 5

so you are doing implicit differentiation in class?

Answer 6

y(x)'=(-2y(x)+19)^0.5 and y(-1)=-3

Answer 7

yes

Answer 8

its implicit

Answer 9

my course is ODE, and we have to use separable equations

Answer 10

that is y'(x)=g(x)h(y) then integral 1/h(y)= integral g(x)dx+c

Answer 11

y(x)'=(-2y(x)+19)^0.5

dy/((-2y+19)^0.5) =dx

Answer 12

and then? I dont not how to integral it , its composite, cause it's not y but y(x)

Answer 13

....

Answer 14

http://i.imgur.com/CMgMz.gif

Answer 15

sec phone

Answer 16

(-1/2)*du/(sqrt(u))

Answer 17

u=-2y+19

go.

Answer 18

u=-2y(x)+19 actually, so du=-2y'(x), this question is not that simple I think

Answer 19

cause I already try to calculate by using ur method before, but I didn't get the right answer

Answer 20

k.

Answer 21

gl with thinking y(x) isn't y.

Answer 22

its a implicit fct, y(x) is a fct about x but not y

Answer 23

you've had a year of calc. ? (T/F)

Answer 24

then I think y'=g(x)h(y) and h(y)=1,g(x)=(-2y(x)+19 )^0.5

Answer 25

T

Answer 26

what did you get for y?

Answer 27

y=integral (-2y(x)+19 )^0.5 dx +C

Answer 28

"cause I already try to calculate by using ur method before, but I didn't get the right answer"

what did you get?

Answer 29

y=-1/3(-2y+19)^1.5

Answer 30

not even close

Answer 31

i am absolutely wrong, and still confused about y(x) and y, so what do you get for y?

Answer 32

integrate the expression I gave you; set it equal to x+C; solve for y.

Answer 33

19/2+(-25/(e^2))(1/2)(e^(-2x))

Answer 34

is that close?

Answer 35

what's the integral of u^(-1/2) ?

Answer 36

thx so much! i finally got the right answer, 3/2-x^2/2+4x, actually at first I got so confused with the f(x), I think its different with y, cause it contains x so cant calculate as y, and finally I found that they r just same, so I got the right answer. Really appreciates a lot !!!!!!