Question

another linear differential equation question, pls help will give medal & fan forevs

Answer 1

@phi

Answer 2

do you really mean the exponent is 2-6 ?

Answer 3

yes that the exponent written in the book

Answer 4

that's the same as r^-4 ??

Answer 5

yes

Answer 6

so what to do now

Answer 7

it looks like a typo and they meant
\[ (4r ^{2}-6) dr +r^3 ds = 0 \]

Answer 8

oops the 6 should be in exponent

Answer 9

\[(4r ^{-4}) dr + r^3 ds = 0\]

Answer 10

whats next

Answer 11

I would multiply the equation by r^-3

Answer 12

oh so it will be \[4r ^{-7} dr + \frac{ 1 }{ ds } = 0\]

Answer 13

then?

Answer 14

how did you flip ds ?

Answer 15

oh just DS only

Answer 16

@phi then what now pls

Answer 17

you can move ds to the other side
\[ 4 r^{-7} \ dr = - ds \]
now integrate both sides

Answer 18

\[\frac{ 4r ^{-8} }{ 8 } = -ds\]

Answer 19

\[\frac{ r ^{-8} }{ 2 } = -ds\] @phi

Answer 20

the right side would become - s + C
you have to do the same thing (i.e. integrate) to both sides to keep it an equation

Answer 21

also, the left side should be negative, right ?

Answer 22

\[\frac{ r ^{-8} }{ 2 } - s = c\]

Answer 23

so I would have done
\[ -\frac{ r ^{-8} }{ 2 } = - s + c\]
or
multiplying by -1:
\[ \frac{ r ^{-8} }{ 2 }= s+ c\]
(-c is still just a constant, all it c)
and finally
\[ \frac{ r ^{-8} }{ 2 } - s = c\]

Answer 24

@phi but the final answer in book is \[s = \frac{ 3 }{ r^2 }+ \frac{ c }{ r^4 }\]

Answer 25

which makes me think the original question has a typo in it.

Answer 26

ohh

Answer 27

Can you take a snapshot of the problem?

Answer 28

okay

Answer 29

it looks like the original problem should be
\[ (4r^{-2} +6 )\ dr + r^3 \ ds =0\]

Answer 30

oh, lets try it

Answer 31

so we divide it by?

Answer 32

yes, divide by r^3, and move ds to the other side

Answer 33

\[(4r + \frac{ 6 }{ r^3 }) dr + ds = 0\]

Answer 34

horribly wrong

Answer 35

4/r