General Solution for Differential Equation: 6x y^2 -6x + 3 y^2 -3

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Homework Help Overview

The discussion revolves around finding the general solution for the differential equation dy/dx = 6x y^2 - 6x + 3 y^2 - 3, which involves concepts related to Bernoulli's equation and separation of variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start solving the differential equation and suggests it may relate to Bernoulli's equation. Other participants discuss rearranging the equation and question whether it can be treated as a simple separation problem.

Discussion Status

Participants are exploring different methods to approach the problem, including Bernoulli's equation and separation of variables. Some guidance has been offered regarding rearranging the equation, but there is still uncertainty about the best method to apply.

Contextual Notes

There is a lack of consensus on the appropriate method to use, with participants questioning the setup and definitions involved in the equation.

intenzxboi
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Give the general solution of:

dy/dx = 6x y^2 -6x + 3 y^2 -3.

No clue how to start. Guessing you would use Bernoulli's equation but i don't see how it is possible to put in the y'(x) + p(x)y(x) = q(x) y(x)^n form.
 
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Bernoulli Equation

\frac{dy}{dx} = f(x)y+g(x)y^k

y^{1-k}=y_1 + y_2

where

\phi(x) = (1-k)\int f(x) dx

y_1 = Ce^\phi

y_2 = (1-k)e^\phi \int e^{-\phi} g(x) dx
 
Actually don't bother, you can rearrange to get this:

\frac{dy}{dx} = (6x+3)(y^2-1)

you can take it from there, no?
 
Yes i can thank you
 
actually I am still having trouble do i still use the bernoulli method?

so it is a simple separation problem?
 
Sorry I didn't realize you replied.

\frac{dy}{dx} = (6x+3)(y^2-1)


\int \frac{dy}{y^2-1} = \int (6x+3) dx


\frac{1}{2}log\left(\frac{1-y}{1+y}\right) = 3x^2+3x + C


\sqrt{\frac{1-y}{1+y}} = Ae^{3x^2+3x}


y=\frac{1-Ae^{6x^2+6x}}{1+Ae^{6x^2+6x}}


Something like this I think?
 
Last edited:

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