Question

Help Pleeeees :3

Answer 1

I need help with Numero Cuatroo. (#4).

Answer 2

@agent0smith

Answer 3

I don't know. Sorry! ♥

Answer 4

Not sure, sorry.

Answer 5

It's okay guys. :/

Answer 6

im sorry idk

Answer 7

what university u go too?

Answer 8

ill hlep u lol xD

Answer 9

first rewrite it in standard form for me
dy/dx + p(x) y = q(x)

Answer 10

518nad always comes in clutch.

Answer 11

@Karim728 I go to CSULB.

Answer 12

Okay, I get... \[\frac{ dy }{ dx } -\frac{ 4 }{ x }y = 6x^3e^x\]

Answer 13

okay so here is how this method works

Answer 14

\[I(x) = e^{\int\limits \frac{ 4 }{ x } dx}\]

Answer 15

Negative*

Answer 16

we try to multiply by some func u(x) such that

y'u(x) + p(x)u(x) y = q(x)u(x)
(Y*u )' = y'*u + y*u' = y' u + p u y = q u
this means u' = p u
so u = e^(int(p) dx)
this u is the integrating factor

Answer 17

Yessir.

Answer 18

if we do this then
(Y*u )' = qu
y= (integral qu dx )/ u

Answer 19

oh dont forget the -4/x

Answer 20

u are doing 4/x

Answer 21

So then
Integrating Factor = x^4
\[\int\limits[x^4y]' = \int\limits 6x^7e^x\]

Answer 22

Oops. forgot the negative. lol

Answer 23

yes

Answer 24

tell me whatu get

Answer 25

\[I(x) = e^{-4 \int\limits \frac{ 1 }{ x }dx} = e^{\ln|x|^{-4}} = x^{-4} = \frac{ 1 }{ x^4 }\]

Thus

\[\int\limits [\frac{ 1 }{ x^4 }y]' = \int\limits \frac{ 6 }{ x } e^x\]

Answer 26

yup

Answer 27

so itsss the last choice

Answer 28

Oh i see. Haha thanks!

Answer 29

hey could you un tag me so i can stop getting notified

Answer 30

ab this quistion

Answer 31

Well that looked difficult...
You two were talking like it was common conversation or something... xP
I was just sitting like "What in the World......"