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

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SUMMARY

The general solution for the differential equation dy/dx = 6x y^2 - 6x + 3 y^2 - 3 is derived using the method of separation of variables. The equation can be rearranged to dy/dx = (6x + 3)(y^2 - 1). By integrating both sides, the solution is found to be y = (1 - Ae^(6x^2 + 6x)) / (1 + Ae^(6x^2 + 6x)), where A is a constant determined by initial conditions. This approach simplifies the problem without needing to apply Bernoulli's equation directly.

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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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