Question

(ODE) trying to solve a prompt, posted below shortly.

Answer 1

\[\text{Solve the differential equation.}\]\[x \frac{dy}{dx}+y(1-3x^3y^2)=0\]

Answer 2

Any help would be appreciated on this.

Answer 3

Do you think I could put the terms on each other side and multiple the differential to get something like an exact equation or something?

Answer 4

@agent0smith

Answer 5

@gleem, should I multiply through by the differentials, or do you have a good idea of how I should attempt to solve this?

Answer 6

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

Answer 7

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

Answer 8

Bernoulli.

Answer 9

the equation can be put in the form\[dy/dx +y/x = 3x ^{2}y ^{2}\] which is a form of Bernoulli's equation.

Answer 10

Yeah, I see it now, thank you.

Answer 11

Alright, moving forward from there:

Answer 12

\[u=y^{1-n}=y^{1-3}=y^{-2}; \ \ \ du = -2y^{-3}y'\]

Answer 13

\[-2y^{-3}\frac{dy}{dx}-\frac{2}{xy^2}=\frac{-6x^2}{y}\]
I did something wrong, I think.

Answer 14

nevermind, I was looking at your equation and you made a mistake; \[-2y^{-3}\frac{dy}{dx}-\frac{2}{xy^2}=-6x^2\]

Answer 15

Okay, now I'm stuck as to what to do.

Answer 16

@Mendicant_Bias please try this substitution:
\[y(x)=\frac{ 1 }{ z(x) }\]

Answer 17

I got this ODE:
\[-\frac{ dz }{ dx }+\frac{ z }{ x }-3x ^{2}=0\]
which can be solved for z(x)
@Mendicant_Bias

Answer 18

How did you know how to do that?

Answer 19

The standard approach would suggest that the substitution should be \[y(x,u) = u/x\] and you should get a DE with variables separated.

Answer 20

That looks something like what's done when you know you're dealing with a homogeneous function, not a Bernoulli DE; Alright, I'll have to sit and think on this, thank you.

Answer 21

The standard approach for BE is the function multiplying y i.e. 1/x should substuted in the integral below as such\[y(x,u) = ue ^{-\int\limits_{}^{}\left( 1/x \right)dx}\] this yields y =u/x

Answer 22

sorry I've made an error befor, I think that this substitution works well:
z(x)=1/y^2