Question

Differential Equation.
Proper substitution method.

Answer 1

\[\frac{ dy }{ dx }= y(xy^3-1)\]

Answer 2

I am currently aware of the y=ux method but it seems the method does not work well

Answer 3

I'm inclined to agree, I can't seem to make any progress with that sub...

This might be one of those problems that require a really tricky substitution. I've tried the following:
\[y=v^{1/3}~~\iff~~y^3=v\\
y'=\frac{1}{3v^{2/3}}v'\]
Subbing into the equation,
\[\begin{align*}\frac{1}{3v^{2/3}}v'&=v^{1/3}\left(vx-1\right)\\
\frac{1}{3}v'&=v(vx-1)
\end{align*}\]
I haven't tried working through it any further, but it looks a bit nicer than before.

Answer 4

How would I arrive at y= v^1/3 myself?

Answer 5

There seems to be another substitution involving x = v y

Answer 6

Well, it was kind of a shot in the dark for me. I had a feeling the \(y^3\) factor was causing all the trouble, so the best way to make it go away is find an expression for \(y\) such that, when you cube it, you can reduce the power.

Also, there's a nice advantage to using rational power substitutions like this one. Notice how all the exponents throughout the equation reduce to an integer?

As for a successive substitution: definitely possible. See where it takes you.

Answer 7

It seems possible with that substitution but I don't want to proceed without knowing how to arrive at the substitution myself but I guess i see it now

Answer 8

If it's not the standard \(y=ux\) substitution, you should examine the equation closely for clues. Keep in mind the derivatives you have to take. Functions like \(e^{f(x)}\) will have derivatives with factors of \(e^{f(x)}\), and similarly \(\sqrt x\) has derivatives containing factors of \(\sqrt x\), so you may be able make terms disappear.

Answer 9

How about distributing the y and setting v = y^-3?

Answer 10

Is that a legal move? I assume any substitution that is consistent works then?

Answer 11

You must be thinking of the Bernoulli substitution. I think it'll work!

Answer 12

What is the Bernoulli substitution? If it helps I would prefer to use it

Answer 13

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

Answer 14

It's the method for solving equations of that form ^^

Answer 15

Doing so will make the equation linear, and from there you find an integrating factor etc.

Answer 16

Alright thank you

Answer 17

You're welcome. I was under the impression you had to come up with some crazy substitution, but I'm glad it's a lot simpler than that.

Answer 18

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

Answer 19

\[V = y^{-3}\]
\[\frac{ dv }{ dx} = -3y\]

Answer 20

Power/chain rule:\[\frac{dv}{dx}=-3y^{-4}\frac{dy}{dx}\]

Answer 21

So
\[-3^{-1}\frac{ dv }{ dx} +v=x\]

Answer 22

\[\frac{ dv }{ dx}-3v=-3x\]

Answer 23

Now this looks familiar,
Integrating factor is
\[e^{-3x}\]
\[e^{-3x}\frac{ dv }{ dx }-3ve^{-3x}=-3xe^{-3x}\]

Answer 24

\[e^{-3}(v)=\int\limits_{}^{}-3x(e^{-3x})\]

Answer 25

Integration by parts?

Answer 26

Yes

Answer 27

Thank you from here I should be fine I appreciate the help

Answer 28

No problem