1. ## Bernoulli's Equation problem

problem:
6y²dx - x(2x³+y)dy = 0

n = ?
P = ?
Q = ?
V = ?

general solution = ?

2. Originally Posted by cazimi
problem:
6y²dx - x(2x³+y)dy = 0

n = ?
P = ?
Q = ?
V = ?

general solution = ?
You first need to re-write this equation in Bernoulli form

$\displaystyle \frac{dy}{dx} + P(x)y = Q(x) y^n$. Isolating $\displaystyle \frac{dy}{dx}$ gives

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

which isn't of the right form but isolating $\displaystyle \frac{dx}{dy}$

gives

$\displaystyle \frac{dx}{dy} - \frac{x}{6y} = \frac{x^4}{3y^2}$ where $\displaystyle P = -\frac{1}{6y}, \;\;\ Q = \frac{1}{3y^2} \,\,\, n = 4$. I don't know what your V is?

Before we discuss the solution, do you know how to solve Bernoulli equations?

3. Yes, I was referring to the V as

$\displaystyle v=exp(\int\P,dx)$

4. Originally Posted by cazimi
Yes, I was referring to the V as

$\displaystyle v=exp(\int\P,dx)$
Before you find your integrating factor, you will need to linearize the equation

so

$\displaystyle \frac{dx}{dy} - \frac{x}{6y} = \frac{x^4}{3y^2}$

becomes

$\displaystyle \frac{1}{x^4} \frac{dx}{dy} - \frac{1}{x^3} \frac{1}{6y} = \frac{1}{3y^2}$

and then let $\displaystyle u = \frac{1}{x^3}$. See how you go from here.