Question

\[y'=y(xy^3-1)\]

Answer 1

first some trivial algebra \[y'=xy^4-y\\
y'+y=xy^4\]

Answer 2

now \(n=?\)

Answer 3

-3

Answer 4

lol, that is the answer to the second question

Answer 5

the first question was \(n=?\) you say "4"
the second question was \(1-n=?\) you say \(-3\)

Answer 6

Lol I understood

Answer 7

\[y'+y=xy^4\] becomes \[u'-3u=-3x\]

Answer 8

Just substituting y = u?

Answer 9

we can go through all the boring steps if you like but remember i said i wanted to go right from here \[y'+p(x)y=g(x)y^n\] to \[u'+(1-n)p(x)u=(1-n)g(x)\]

Answer 10

Yes

Answer 11

since n was 4, 1-n was -3 i went right to it

Answer 12

we can take all the boring steps, but it seems pointless
if it works in general it has to work for a specific example

Answer 13

Pleaseeee

Answer 14

ok but of course we are not done, we have to still solve the equation right?

Answer 15

Yes

Answer 16

\[u'-3u=-3x\] integrating factor is \(e^{-3}\)

Answer 17

you get \[e^{-3x}u'-3ue^{-3x}=-3xe^{-3x}\]

Answer 18

Where did you get the integrating factor?

Answer 19

i made a mistake

Answer 20

\[p(x)=-3\\
\int -3dx=-3x\\
e^{-3x}\]

Answer 21

this was right though\[e^{-3x}u'-3ue^{-3x}=-3xe^{-3x}\]

Answer 22

Wait

Answer 23

Where is the work for u ^-3 = y

Answer 24

grrr

Answer 25

Pleaseeeee

Answer 26

ok dammit

Answer 27

\[y'+y=xy^4\]divide by \(y^4\) get \[y^{-4}y'+y^{-3}=x\]

Answer 28

put \(u=y^{-3}\) so \(u'=-3y^{-4}\) or \(-\frac{1}{3}u'=y^{-4}\)

Answer 29

Ok

Answer 30

get \[-\frac{1}{3}u'+u=x\] or \[u'-3u=-3x\]

Answer 31

Cool

Answer 32

\[y'+y=xy^4\] becomes \[u'-3u=-3x\]

Answer 33

Wait

Answer 34

there is one tiny sophistication but it works \[\frac{dy}{dx}\] is what you are given
it is the same as \[\frac{du}{dy}\frac{du}{dx}\] i skipped wring that

Answer 35

Dumb question but if u' = -3y^-4 and -1/3u'=y^-4

Answer 36

oops\[\frac{du}{dy}\frac{dy}{dx}\]

Answer 37

why would it be u'

Answer 38

that is the part i skipped lets do it slowly

Answer 39

Sorry

Answer 40

\[y^{-4}y'+y^{-3}=x\] which we better write as \[y^{-4}\frac{dy}{dx}+y^{-3}=x\]

Answer 41

\[u=y^{-3}\\
\frac{du}{dy}=-3y^{-4}\\
-\frac{1}{3}\frac{du}{dy}=y^{-4}\]

Answer 42

\[\huge \color{red}{y^{-4}}\frac{dy}{dx}+y^{-3}=x\]
\[\huge \color{red}{-\frac{1}{3}\frac{du}{dy}}\frac{dy}{dx}+u=x\]

Answer 43

so \[-\frac{1}{3}\frac{du}{dx}+u=x\\
\frac{du}{dx}-3u=-3x\]

Answer 44

i guess i can't convince you you don't need these steps

Answer 45

Ok I get it

Answer 46

we still did not solve 17, just went from \[y'+y=xy^4\] to \[u'-3u=-3x\] in too many steps

Answer 47

I'm sorry!

Answer 48

i am just trying to convince you that it is easy
it is good to understand why it works though, so a worked out example is fine but it is real real simple to change the form

Answer 49

Let's do the next one in a shortcut. I just have to see the long way first you know that.

Answer 50

\[y'+p(x)y=f(x)y^n\\
u'+(1-n)p(x)u=(1-n)g(x)\]

Answer 51

\[u'-3u=-3x\] you can do need integration by parts

Answer 52

\[e^{-3x}u'-3ue^{-3x}=-3xe^{-3x}\] then\[ \frac{d}{dx}[ue^{-3x}]=-3xe^{-3x}\] then \[ue^{-3x}=\int -3xe^{-3x}dx\]

Answer 53

cheat to do the integration

Answer 54

I love cheating

Answer 55

\[ue^{-3x}=xe^{-3x}+e^{-3x}+c\]

Answer 56

nope

Answer 57

\[ue^{-3x}=xe^{-3x}+\frac{1}{3}e^{-3x}+c\]

Answer 58

multiply by \(e^{3x}\) so solve for \(u\) get \

Answer 59

What's the gimmick?

Answer 60

to integrate? parts

Answer 61

\[ue^{-3x}=xe^{-3x}+\frac{1}{3}e^{-3x}+c\]
\

Answer 62

dammit

Answer 63

latex isn't working again

Answer 64

\

Answer 65

Oh my.