Question

need help with second derivatives of calculus problems

Answer 1

\[2y-x+xy=4\]

Answer 2

\[\frac{ dy }{ dx }=\frac{ 1-y }{ 2+x }\] I know this is the first derivative

Answer 3

I need help finding the second derivative

Answer 4

so, apply quotient rule

Answer 5

or if you want to simplify things,

Answer 6

2y' -1 +xy' +y = 0
take the derivative here itself!

Answer 7

ok give me a second, nice to see you again, I have been very busy lately that is why I have not been around

Answer 8

no problem, good to see you too :)

Answer 9

i would suggest applying derivative here 2y' -1 +xy' +y = 0
instead of quotient rule...

Answer 10

yes I was using dy/dx notation and it was getting messy

Answer 11

2y"+y'+y"+y'=0 correct

Answer 12

3y"+2y'=0

Answer 13

I sub for y' correct

Answer 14

wait, how did you differentiate xy' ?

Answer 15

who knows I don't really use prime notation on these, good chance I messed it up

Answer 16

ok,
for xy'
first we diff. x and keep y' as it is
so, just y'
then we diff. y' and keep x as it is
so, xy''

so, derivative of xy' will be y' + xy''
got this ?

Answer 17

so it should be 2y"+y'+y"x+y'=0

Answer 18

2y"+y'+xy"+y' = 0
same thing , right ?

Answer 19

yes, but did you get how ?

Answer 20

yes sorry looking at my paper
so y"=(-2y')/(2+x) now I just sub for y' correct

Answer 21

yes, correct

Answer 22

cool thank you. I will try to switch notation (makes the problem easier to deal with)

Answer 23

welcome ^_^