Question

can someone help me with this one....
eliminate the arbitary constant of
x^2 + y^2 - Ay =B
the answer is xy`` - (y`)^3 - y` = 0

Answer 1

\[equation : x ^{2} + y ^{2} -Ay = B\]

Answer 2

\[final answer: xy ^{``} - (y ^{`})^{3} - y ^{`} = 0\]

Answer 3

@rvc

Answer 4

@SamsungFanBoy @zzr0ck3r @across

Answer 5

@baru

Answer 6

can you differentiate the given equation?

Answer 7

\[x^2 + y^2 - Ay = B\] @baru

Answer 8

yes, differentiate that with respect to x

Answer 9

how?

Answer 10

\[\frac{d(x^2 + y^2 - Ay}{dx} = \frac{d (B)}{dx}\]

Answer 11

\[2x + 2yy'-Ay' = 0\]

Answer 12

ah. thats what you mean by differentiate

Answer 13

okay got that part

Answer 14

\[2x + 2yy'-Ay' = 0\]

can you differentiate this again?

Answer 15

\[d(2x + 2yy`- Ay`)/dx = 0\]

Answer 16

yes,
what so what do you get after you carry out the differentiation

Answer 17

\[2 + y` - A = 0\]

Answer 18

no,
y is a function of x
you have to differentiate with respect to x

\[dy/dx=y'\\d(y')/dx=y''\]

Answer 19

ah, so 2y + 2yy`` - A yy`

Answer 20

\[d(2x + 2yy`- Ay`)/dx = 0 \\=2 + 2yy'' +2y'y' - Ay''\]

Answer 21

why Ay` become 2y`y1?

Answer 22

no.. i mean why theres a 2y`y`?

Answer 23

its called the 'product rule' of differentiation\[\frac{d(2yy')}{dx} = 2y \frac{d(y')}{dx} + y' \frac{d(2y)}{dx}\]

Answer 24

didnt know about that. whats the formula for that?

Answer 25

if u and v are functions of x\[\frac{d(uv)}{dx}=u \frac{dv}{dx} + v \frac{du}{dx}\]

Answer 26

ah, so you use the power rule on 2yy`?

Answer 27

i used the 'product rule'

Answer 28

yes the product rule i mean

Answer 29

so what next

Answer 30

\[x^2 + y^2 - Ay = B\\ 2x + 2yy'-Ay' = 0\\ 2 + 2yy'' +2y'y' - Ay''=0\]

now we need you use these three equations to some how get rid of A and B

Answer 31

re-arrange the second equation
\[2x + 2yy'-Ay' = 0\\2x + 2yy' = Ay'\\A=\frac{2x}{y'} + 2y\]

Answer 32

substitute A in the third equation

Answer 33

\[2 + 2yy'' +2y'y' - ( \frac{2x}{y'}+2y)y''=0\]

Answer 34

yes i got the same thing

Answer 35

so y` and the y in 2y will be cancel?

Answer 36

which part are you talking about

Answer 37

the substituting the A

Answer 38

i just divided both sides by y'

Answer 39

ah, okay got that one

Answer 40

so will going to subtract all the equations?

Answer 41

no,
we substituted A and thus eliminated it
\[2 + 2yy'' +2y'y' - ( \frac{2x}{y'}+2y)y''=0\]

all thats left is to simplify the above equation so that it matches the given answer

Answer 42

but the answer is \[xy` -(y`)^3 - y` = 0\]

Answer 43

as i said, we need to simplify the above equation

so just multiply both sides by \(\frac{y'}{2}\)

Answer 44

do that and you should have your answer

Answer 45

all the side by 2 so.. \[yy` +y`y` (x/y` +y) y` = 0\]

Answer 46

\[(2 + 2yy'' +2y'y' - ( \frac{2x}{y'}+2y)y'')) \times \frac{y'}{2}=0 \times \frac{y'}{2}\]

Answer 47

why u didnt divide the whole equation by 2?

Answer 48

so y` + y`` + y` - x y`

Answer 49

i get
\[y' + yy'y'' +(y')^3 -xy''-yy'y''=0\]

simplify further
\[y' +(y')^3 -xy''=0\]

multiply both sides by -1 and rearrange

\[xy'' - (y')^3 -y' =0\]

Answer 50

ah wait why you multipley all sides by y`/2?

Answer 51

i did that to simplify the expression