Question

How do I solve the differential equation x^2+(x^3+8)y'=0

Answer 1

Can I divide each term by 1/dy to separate?

Answer 2

This is in the form m+ny'=0

Answer 3

see if its exact

Answer 4

do the partial with respect to y for the m and then do the partial with respect to x in the n portion

Answer 5

Start by isolating y' .. then integrate both sides

Answer 6

if they are equal then it is exact

Answer 7

But to isolate y' I would need to divide each term by y' correct?

Answer 8

wait...is this cal 1 or differential equations and linear algebra

Answer 9

Calc II differential equations

Answer 10

http://1337.is/~gaulzi/tex2png/view.php?png=201103312034348813.png would look like this isolated

Answer 11

Well you can try implicit differentiation.

Answer 12

take the derivative of the first term with respect to x

Answer 13

kristin wouldn't it be -x^2?

Answer 14

sorry yes, thats correct, minor typo :)

Answer 15

Ok no problem. Thanks

Answer 16

I can toss up for you the answer if you want :)

Answer 17

y' is dy/dx correct? If i were to rewrite it

Answer 18

yeah

Answer 19

Ok so if I were to solve y'+y=10 would I just replace y' with dy/dx and then move it all around?

Answer 20

I'm trying to figure out the steps

Answer 21

it can be done by the separable variables method

Answer 22

after some modification you can get
dy=-x^2/(x^3+8) dx
just integrate both sides

Answer 23

http://1337.is/~gaulzi/tex2png/view.php?png=201103312042017465.png these are the two answers you can get.. depends if you like to use ln or log

Answer 24

I know the answer I'm trying to figure out the steps, could you show me the steps kristin? Plz.

Answer 25

I can show it to you

Answer 26

did you integrate like she said?

Answer 27

I know I need to integrate each piece, I just need to know how to get each piece by itself

Answer 28

do a u substitution

Answer 29

u=x^3+8
du=3xdu

Answer 30

du=3x^2 sorry

Answer 31

you should get: integral( -1/3(1/u) dx

Answer 32

just take u=x^3+8 --> du=3x^2dx
substitute in the integral you will get
\[y=-1/3\int\limits_{}^{}(1/u)du=-1/3\ln \left| u \right|\]

Answer 33

now just substitute for u=x^3+8
\[y=-1/3\ln \left| x^3+8 \right|+c\]

Answer 34

Oh right I got that. Thanks

Answer 35

np