Question

1

Answer 1

start by dividing xy^2 thru out

Answer 2

-(y'x)/y^2 - 1/(xy) =1 ?

Answer 3

you should get :

\(\large \mathbb{ \frac{y'}{y^2} +\frac{1}{xy} = -1}\)

Answer 4

oh ya okay,why leave out the minus?

Answer 5

lol dint notice that much.. .

next, convert it to linear by substituting : \(\large v = \frac{1}{y}\)

Answer 6

okay, question, the equation is like this: dy/dx +P(x)y=Q(x)y^n, P(x)=1/x ; Q(x)=-1 , v=1/y taking the variable of P(x) isit?

Answer 7

i mean the 2nd term on the LHS.

Answer 8

ok, lets put it in standard bernouli form first... so it becomes easy to see

Answer 9

\(\large \mathbb{ xy'+y=-xy^2}\)

divide x both sides

\(\large \mathbb{ y'+\frac{1}{x}y=-y^2}\)

Answer 10

now compare it wid the bernouli standard equation you're having..

Answer 11

P(x)=1/x, Q(x)=-1, n=2, i want to know how v=1/y

Answer 12

\(\large v = \frac{1}{y^{n-1}}\)

since n = 2,

\(\large v = \frac{1}{y^{2-1}} = v = \frac{1}{y} \)

Answer 13

u will see how it helps in converting the existing equation to a good linear equation shortly..

Answer 14

ohh i see, i almost forgot abt tht part, the u=y(1-n)

Answer 15

but before that, divide y^2 both sides...

Answer 16

so y'/y^2 - 1/xy = 1

Answer 17

\(\large \mathbb{ xy'+y=-xy^2} \)

divide x both sides

\(\large \mathbb{ y'+\frac{1}{x}y=-y^2} \)

divide y^2 both sides

\(\large \mathbb{ \frac{y'}{y^2} +\frac{1}{xy} = -1}\)

substitute \(\large \mathbb{v=\frac{1}{y} \implies v' = \frac{-y'}{y^2}}\)

Answer 18

\(\large \mathbb{ -v' +\frac{1}{x}v = -1}\)

Answer 19

this is ur familiar linear equation in v's which u can solve right ?

Answer 20

wait, isnt v'= -1/y^2 why isit -y'/y^2?

Answer 21

dont forget, we're differentiating w.r.t x

Answer 22

dv/dx = f(y)' * dy/dx

Answer 23

v' = -1/y^2 * y'

Answer 24

y is a function of x ok ?

Answer 25

ahhh okay got it(:

Answer 26

good, see if u can solve the linear equation

Answer 27

A Bernoulli equation is of the form
\[y'+p(x)y=q(x)y^n\]
With the substitution \[v=y^{1-n}\]\[v'=(1-n)y^{-n}y'\]
\[y'=\frac1{1-n}y^nv'\]
\[\frac1{1-n}y^nv'+p(x)y=q(x)y^n\]\[\frac1{1-n}v'+p(x)y^{1-n}=q(x)\]
The equation is reduced to the linear DE
\[v'+(1-n)p(x)v=(1-n)q(x)\]

Answer 28

got it! thank you! (:

Answer 29

cool :)

Answer 30

why have you changed the title of this question @ememlove ?

Answer 31

oh because its an exam question from my sch exam question bank.. i just dont want to get in trouble since i quote "Publication, distribution, mount on any electronic network, or retaining portions of licensed materials or combining them with any other material is prohibited." its better to change the name so that it wont be in top searches if ever. haha just precautions, i still had to do it since i want to learn.

Answer 32

EXAM‽

Answer 33

yeah.. its like past year exam papers database. students can dl them but they are not allowed to do some things with it. like what i quote