Question

\[\frac{dy}{dx}-\frac{2y}{x}=\frac{3y^4}{x^2}\]

Answer 1

this is where we are, right?

Answer 2

👍

Answer 3

ok now if the y part was not on the right, this would be one like from 2.3, linear

Answer 4

So whats the difference between bernoulli and linear?

Answer 5

\[\huge \frac{dy}{dx}-\frac{2y}{x}=\frac{3\color{red}{y^4}}{x^2}\]

Answer 6

you gotta get rid of that \(y^4\) to make it linear. It is bernoulli because that \(y^4\) is there

Answer 7

Does y have to be on both sides as well to make it a bernoulli? Or can it be x = y^4?

Answer 8

\[y'+p(x)y=f(x)\] is linear
\[y'+p(x)y=f(x)y^n\]is bernoulli

Answer 9

Oh! So linear doesnt have any y on the right side?

Answer 10

no it does not

Answer 11

2.3 had definition

Answer 12

That makes so much more sense.

Answer 13

This textbook sucks. I actually miss stewarts now lol

Answer 14

so gimmick is to make bernoulli linear by making a substitution

Answer 15

you make the magic substitution \ and by some miracle it goes

Answer 16

u = y/x ?

Answer 17

\[\huge u=y^{1-n}\]

Answer 18

Oops

Answer 19

that was for the other type

Answer 20

Ok

Answer 21

homo genius

Answer 22

so i guess you missed why the magic trick works right?

Answer 23

Yes

Answer 24

ok we can see what happens by doing it in this example, although 15 was easier (kinda)

Answer 25

Okie

Answer 26

\[\huge \frac{dy}{dx}-\frac{2y}{x}=\frac{3\color{red}{y^4}}{x^2}\]evidently \(n=4\) so substitution is ...

Answer 27

\(n=4\)so substitution is ?

Answer 28

Y^-3

Answer 29

yeah \(u=y^{-3}\) first step is to solve that for \(y\) and then calculate \(\frac{dy}{du}\)

Answer 30

How?

Answer 31

are you asking me how to solve \(u=y^{-3}\) for \(y\)??

Answer 32

No I thought to solve for y’ it’s the chain rule

Answer 33

no, first solve for y, then compute \(\frac{dy}{du}\) will make your life much easier before we do any subsitution

Answer 34

Ok

Answer 35

well...

Answer 36

Idk

Answer 37

jeez i have to do everything \[u=y^{-3}\\
\huge u^{-\frac{1}{3}}=y\]

Answer 38

I thought so. I just forgot my minus sign

Answer 39

now what is \(\frac{dy}{du}\) ?

Answer 40

U^(-1/3)

Answer 41

no, that is just y

Answer 42

take the derivative yo

Answer 43

0?

Answer 44

better yet, call me, this is taking more time than it is worth

Answer 45

\

Answer 46

\

Answer 47

\(y=u^{-\frac{1}{3}}\)
\(\frac{dy}{du}=-\frac{1}{3}u^{-\frac{1}{3}-1}\)