Question

Can someone help me with some Differential Equation?

Answer 1

\[y'+x*\sqrt[3]{y}=3y\]

Answer 2

@perl

Answer 3

@rvc

Answer 4

it's not easy

Answer 5

$$
\Large{
y'+x \cdot \sqrt[3]{y}=3y
\\ y'+x \cdot {y}^{1/3}=3y


}
$$

Answer 6

yeah I did that, so we get a linear eqation

Answer 7

I did it through the variation method y=UV

Answer 8

that is not a linear de , though

Answer 9

it's not? why?

Answer 10

because of the power of 1/3

Answer 11

on the y

Answer 12

oh

Answer 13

only powers of 1 on y and its derivatives in a linear d.e.

Therefore, this is a linear d.e.
y ' + p(x) y = g(x)

Answer 14

ok can you show me your work on variation

Answer 15

what do you mean?

Answer 16

Hipocampus
I did it through the variation method y=UV

Answer 17

\[U'V+UV'-3UV=-X(UV^\frac{ 1 }{ 3 }\]

Answer 18

now, I would take the first and the third parts.and equal them to zero, right?

Answer 19

so \[V(U'-3U)+UV'=X(UV)^\frac{ 1 }{ 3 }\]

Answer 20

looks good

Answer 21

now \[U'\]

Answer 22

\[U'-3U=0\]

Answer 23

to zero

Answer 24

right?

Answer 25

@per?

Answer 26

@perl

Answer 27

I got \[U=e^(3x)\]

Answer 28

and from here it's not working

Answer 29

@freckles can you help me from here? now I compare \[UV'=X(UV)^\frac{ 1 }{ 3 }\]

Answer 30

@perl ? Did you quit?

Answer 31

@shifuyanli

Answer 32

@Loser66 HELP!

Answer 33

@phi ! will you be my savior?

Answer 34

I don't know

Answer 35

NO!

Answer 36

@rational

Answer 37

but @phi ? You know EVERYTHING!

Answer 38

idk what to do about it

Answer 39

let\[z^n=y\]
\[nz^{n-1}z'=y'\]
\[y'+xy^{k}=3y\]
\[nz^{n-1}z'+x(z^n)^{k}=3z^n\]
\[z'+\frac xn\frac{z^{kn}}{z^{n-1}}=\frac{3z^n}{nz^{n-1}}\]
what does n have to be to get rid of that offending part?

Answer 40

kn-n+1 = 0, solve for n

Answer 41

wait from where do you start @phi?

Answer 42

I think amistre is showing how to do it.

Answer 43

\[z'+\frac xn\color{red}{\frac{z^{kn}}{z^{n-1}}}=\frac{3z^n}{nz^{n-1}}\]
\[z'+\frac xn\color{red}{\frac{z^{kn}}{z^{n-1}}}=\frac{3}{n}z\]
\[set~\frac{z^{kn}}{z^{n-1}}=1\]

Answer 44

oh my, we didn't learn that

Answer 45

baynes ... or something cant recall the name of it. but it helps convert non linear into linear

Answer 46

I think we've learned smth similar, I gtg to work but thanks a lot you all!

Answer 47

good luck

Answer 48

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

burn-noodle .... well i had the B part right lol