Question

given y>0 and dy/dx=3x^2+4x/y. if the point (1,sqrt{10}) is on the graph relating x and y, then what is y when x=0?

Answer 1

do you know to find y given y'?

Answer 2

dy/dx

Answer 3

?

Answer 4

you have to integrate y' to get y

Answer 5

oh wait

Answer 6

i didn't see that other y

Answer 7

is it \[y'=\frac{3x^2+4x}{y} ?\]

Answer 8

yes

Answer 9

ok do you know separation of variables?

Answer 10

quotient rule product rule ?

Answer 11

\[\frac{dy}{dx}=\frac{3x^2+4x}{y}\]
you want y so we need to integrate but before we do that we need to separate our y's from our x's

Answer 12

\[y dy=(3x^2+4x) dx\]

Answer 13

oh yes i know how to do that

Answer 14

i multiplied y on both sides and I multiplied dx on both sides
now we need to integrate both sides

Answer 15

\[\int\limits_{}^{}y dy=\int\limits_{}^{}(3x^2+4x) dx\]

Answer 16

whats that long thing u just put

Answer 17

\[\frac{y^2}{2}=\frac{3x^3}{3}+\frac{4x^2}{2}+C\]
whenever you integrate you need to do +C

Answer 18

i havent learned that thing yet

Answer 19

its the only way

Answer 20

what about finding the antiderivative?

Answer 21

do you know what that means?

Answer 22

i learned that friday

Answer 23

yes

Answer 24

thats with weird symbol means

Answer 25

what*

Answer 26

but i dont think i can use that symbol yet

Answer 27

hmmm... that's weird to me

Answer 28

oh :/

Answer 29

its fine i guess i should skip that problem and just ask the teacher 2morrow. ill put up a new one.

Answer 30

ok but that is all i did was found the antiderivative of both sides with respect to whatever varaible was in d( )

Answer 31

oh okay thanks either way :)

Answer 32

np