Question

linear algebra & differential equation question
using bernoulli equation
y'-2(sinx)y = -2y^3/2sinx

Answer 1

it's already in the bernoulli form...so now..identify what y^n is

Answer 2

how far along are you?

Answer 3

um.. y^n would be y^3/2?

Answer 4

I think it's supposed to be y^3 in your problem statement.

Answer 5

? what do you mean \[y'-2(sinx)y = -2y ^{3/2}sinx\]

Answer 6

i dont know what to do after

Answer 7

ok just making sure..

Answer 8

divide through by y^(3/2)

Answer 9

make the sub.s y^(-1/2) =v

Answer 10

you there?

Answer 11

yes

Answer 12

ok... now find v' , remember to use the chain rule...

Answer 13

wait where did you get y^1/2?

Answer 14

v= y^(1-n) n is 3/2

Answer 15

is that what you're asking about?

Answer 16

yes oh isee you divide all of them with y^3/2

Answer 17

cool. So what did you get for v'?

Answer 18

i got y^-3/2 - 2(sinx)y^-v = -2sinx

Answer 19

y' on the y^3/2

Answer 20

v' = -1/2 * y^(-3/2) * y'

Answer 21

now make the sub.s so all the y and y' are in terms of v and v'

Answer 22

\[y ^{-3/2}y' -2(sinx)y ^{-1/2} = -2sinx\]

Answer 23

this is my step right now

Answer 24

y^(-1/2) = v
y^(-3/2) * y' = -2*v'

Answer 25

now you have an equation in v, v' and x ...
you can solve for v using the method of 'integrating factor' or whatever else you are comfortable with.

Answer 26

so you changed the equation into this? what happened to the sinx?

Answer 27

nothing

Answer 28

can you show me the full equation what you changed im not quite getting it..

Answer 29

show me what you have...

Answer 30

\[y ^{-3/2}y' -2(sinx)y ^{-1/2} = -2sinx\]

Answer 31

y^(-1/2) = v
y^(-3/2) * y' = -2*v'

Answer 32

so it would be -2v' - 2(sinx)v = -2sinx ??

Answer 33

v' +(sinx)v = sinx

Answer 34

oh isee then use the 1st order to solve? or homogenous?

Answer 35

"backspace" is sticking lol

Answer 36

yep:)

Answer 37

that is confusing about bernoulli is am i allow to just change y^(-3/2)y' into just v' ? and how did we get the 2v' i get the v' part but how did we get the 2?

Answer 38

how did you know to make it y^(-3/2) * y' = -2*v'

Answer 39

-2? from

Answer 40

y^(-1/2) =v
-1/2 * y^(-3/2) * y' = v'

Answer 41

just took the derivative of the "v" we chose...

Answer 42

all good?

Answer 43

yes i think i got it we took the derivative because its v' right?

Answer 44

we took the derivative because we're trying to substitute in a simpler expression for y' and any (and all) powers of y. Can't do that unless we have v and v'.

Answer 45

one last question can it be different variable? or it has to be v and v'?

Answer 46

It can be whatever you like as long as it's not x :)

Answer 47

thank you~~~~~

Answer 48

Sure:)