Bernoulli's Equation problem

**cazimi** · January 20th 2009, 09:27 PM

problem:
6y²dx - x(2x³+y)dy = 0

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

general solution = ?

**Danny** · January 21st 2009, 04:30 AM

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

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

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

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

gives

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

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

**cazimi** · January 21st 2009, 05:03 AM

Yes, I was referring to the V as

$<br /> v=exp(\int\P,dx)$

**Danny** · January 21st 2009, 09:07 AM

Originally Posted by cazimi

Yes, I was referring to the V as

$<br /> v=exp(\int\P,dx)$

Before you find your integrating factor, you will need to linearize the equation

so

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

becomes

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

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

Thread: Bernoulli's Equation problem

LinkBack

Thread Tools

Display

Bernoulli's Equation problem

Similar Math Help Forum Discussions

I have a problem solving this Bernoulli equation (tried several times)

Bernoulli's Equation - Fluid Flow Problem

Use of Bernoulli's Equation

Bernoulli equation

bernoulli equation

Search Tags