Question

\[y'+p(x)y=g(x)y^n\\
y^{-n}y'+p(x)y^{1-n}=g(x)\]
pot \(u=y^{1-n}\) so \(\frac{du}{dy}=(1-n)y^{-n}\) or \(y^{-n}=\frac{1}{1-n}\frac{du}{dy}\)
substitute back gives \[\frac{1}{1-n}\frac{du}{dx}+p(x)u=g(x)\] good to go
or best is \[\frac{du}{dx}+(1-n)p(x)u=(1-n)g(x)\]

Answer 1

Hello! Sorry I'm late

Answer 2

no problem
any questions about any steps? really you should go right from the top to the bottom, the intermediate steps are not necessary when you are solving one of these

Answer 3

btw i noticed there is no explanation in the book, so i kind of made one up on my own. i like this explanation better

Answer 4

I appreciate it. I'm just going to write this down give me one second.

Answer 5

ok i think i only skipped one step or one part of the explanation

Answer 6

I thought it was n/1-n

Answer 7

it is if you solve back for y

Answer 8

but i realized it is not necessary, think instead of dividing both sides by \(y^n\)as a first step

Answer 9

\[y'+p(x)y=g(x)y^n\] this is your basic bernoulli

Answer 10

\[y^{-n}y'+p(x)y^{1-n}=g(x)\] i mean

Answer 11

Okie

Answer 12

\[\huge y^{-n}y'+p(x)\overbrace{\color{red}{y^{1-n}}}^{\text{ this is your u}}=g(x)\]

Answer 13

Right

Answer 14

then \[\frac{du}{dy}=(1-n)y^{-n}\] yes?

Answer 15

Yes

Answer 16

so \[\frac{1}{1-n}\frac{du}{dy}=y^{-n}\] yes?

Answer 17

Yes

Answer 18

\[\huge\overbrace{ y^{-n}}^{\text{put that here}}y'+p(x)y^{1-n}=g(x)\]

Answer 19

OH! Its just a bunch of subsitution

Answer 20

\[\frac{1}{1-n}\frac{du}{dy}\frac{dy}{dx}+p(x)u=g(x)\]

Answer 21

note that \[\frac{du}{dy}\frac{dy}{dx}=\frac{du}{dx}\]so we get \[\frac{1}{1-n}\frac{du}{dx}+p(x)y=g(x)\]

Answer 22

muliiply by \(1-n\) to get \[u'+(1-n)p(x)=(1-n)q(x)\]

Answer 23

linear

Answer 24

want to do an example? one where \(n\) is not \(2\) which seems to be the most common, maybe on your test but we should try a different one

Answer 25

So is (du/dy) is (1/1-n)?

Answer 26

no

Answer 27

Please let's do an example

Answer 28

\[u=y^{1-n}\\
\frac{du}{dy}=(1-n)y^{-n}\]

Answer 29

let me know when that is clear

Answer 30

\[u=y^4\\
u'=4y^3\] for example

Answer 31

which question do you want to do?

Answer 32

Whatever you choose.

Answer 33

17?

Answer 34

Sure

Answer 35

ill make a new thread, but i want to go right from here \[y'+p(x)y=g(x)y^n\] to here \[u'+(1-n)p(x)=(1-n)q(x)\]

Answer 36

Cool