Question

answered question

Answer 1

bernoulli's Equation

Answer 2

what do you get if you re arrange to

dy/dx=

Answer 3

sorry but i dont know .................it equal to 1 dont know if I am correct >.<

Answer 4

after your get the left hand side to be dy/dx
sub v=y/x

Answer 5

\[\frac{ xdy }{ y-xdx }=1\]

Answer 6

its right?

Answer 7

yeah, now move the other terms to the other side

Answer 8

\[\frac{ dy }{ dx }= \frac{ y-x }{ x }\]

Answer 9

Im correct?

Answer 10

good

Answer 11

then?

Answer 12

let v

Answer 13

how?

Answer 14

\[(y-x)/x = y/x - x/x\]

Answer 15

yes

Answer 16

after that

Answer 17

x/x simplifies to ...

Answer 18

1

Answer 19

good so you have \[\frac{dy}{dx}=\frac yx-1\]


now let \(y/x=v\)

\(y=vx\)

\(y'=?\)

Answer 20

use the product rule to find y'

Answer 21

\[y'=v(1) -x\frac{dv }{ dx }\]

Answer 22

should be plus not minus

Answer 23

ay sorry about that

Answer 24

y′=v(1)+xdv/dx

Answer 25

now sub this in
\[\frac{dy}{dx}=y/x-1\]
becomes
\[v+x\frac{dv}{dx}=v-1\]

Answer 26

which is now variables separable

Answer 27

\[\frac{ dv }{ dx }= \frac{ v ^{2} -v}{ x }\]

Answer 28

um you should have taken away the v from both sides, before dividing by x

Answer 29

$$(x-y)\,dx+x\,dy=0$$note this is homogeneous; both our functions \(x-y,x\) are homogeneous. as @UnkleRhaukus stated, use a substitution \(y=vx\) i.e. \(dy=\left(x\dfrac{dv}{dx}+v\right)\,dx\):$$(x-vx)dx+x(x\,dv+v\,dx)=0\\(x-vx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=-x\,dx\\dv=-\frac1x\,dx\\v=-\log x+C\\y/x=-\log x+C\\y=-x\log x+Cx$$

Answer 30

http://en.wikipedia.org/wiki/Homogeneous_differential_equation#Homogeneous_functions

Answer 31

guys can be this one \[xdv=(v ^{2}-v) dx .....then....... (v ^{2}-v)-xdv=0\]

Answer 32

?

Answer 33

then identify if its exact DE or not

Answer 34

?????

Answer 35

\[v+x\frac{dv}{dx}=v-1\\
x\frac{dv}{dx}=-1\\dv=-xdx\\
\\∫dv=-∫dx/x\]

Answer 36

What is the second thing? where does the dx go?

Answer 37

v=- lnx +c is that right?

Answer 38

yes, now remember v=y/x

Answer 39

then sub v= y/x to v =-ln x +c

Answer 40

y = -xlnx+c ?

Answer 41

almost,

Answer 42

how ?

Answer 43

the last term,

Answer 44

cx

Answer 45

yes

Answer 46

sorry about that

Answer 47

thats better \(\checkmark\)

Answer 48

tnx i other one here tech me ?

Answer 49

xdx + (y-2x )dy

Answer 50

is there an equals sign somewhere?

Answer 51

=0

Answer 52

\[\frac{ dy }{ dx }=\frac{ -x }{ y- }\]

Answer 53

2x

Answer 54

ok this one is the same technique, rearrange to dy/dx=

and substitute v=y/x

Answer 55

divide both sides of the fraction by x, to get it in the right form \[\frac{-x}{y-2x}=\frac{-x/x}{y/x-2x/x }\]

Answer 56

always? v=y/x even to other problem?

Answer 57

When we can express the DE as

y'(x) = f (y/x)

we can call this a homogenous equation
and v = y/x substitution will work

if we can't get this form y'(x) = f (y/x)

this technique wont work

Answer 58

ah i see

Answer 59

can you gve to me the final answer here this will serve as my basis

Answer 60

sometimes a 'homogenous equations" means something else

Answer 61

you gve a sense to me :)

Answer 62

i haven't worked out the final answer yet

Answer 63

by the way thank you :))) tnz alot