# Bernoulli's Equation

• Sep 5th 2011, 06:33 AM
Mike9182
Bernoulli's Equation
What did I do wrong here?

$\displaystyle \frac{dy}{dx} + \frac{2}{x}y = y^2$

$\displaystyle \Rightarrow (-1)y'y^{-2}+(-1)y^{-1}$

$\displaystyle \Rightarrow (-1)y^{-2}y'+(-1)y^{-1}=-1$

$\displaystyle \Rightarrow y^{-2}y'+y^{-1}=1$

$\displaystyle z = y^{-1}$
$\displaystyle \frac{dz}{dx} = -y^{-2}\frac{dy}{dex}$

$\displaystyle \Rightarrow -\frac{dz}{dx}+z=1$

$\displaystyle \Rightarrow \frac{dz}{dx}+(-1)z=-1$

$\displaystyle I = e^{\int{-1 dx}} = e^{-x}$

$\displaystyle e^{-x}z = \int{-1 dx}$

$\displaystyle \Rightarrow e^{-x}z = -x + C$

$\displaystyle \Rightarrow z = e^{x}(-x + C)$

$\displaystyle \Rightarrow y^{-1} = e^{x}(-x + C)$

$\displaystyle \Rightarrow \frac{1}{e^{x}(-x + C)}$

The correct answer given by the professor is:
$\displaystyle \frac{1}{1+Cx}$
• Sep 5th 2011, 07:29 AM
chisigma
Re: Bernoulli's Equation
Quote:

Originally Posted by Mike9182
What did I do wrong here?

$\displaystyle \frac{dy}{dx} + \frac{2}{x}y = y^2$

$\displaystyle \Rightarrow (-1)y'y^{-2}+(-1)y^{-1}$

$\displaystyle \Rightarrow (-1)y^{-2}y'+(-1)y^{-1}=-1$

$\displaystyle \Rightarrow y^{-2}y'+y^{-1}=1$

$\displaystyle z = y^{-1}$
$\displaystyle \frac{dz}{dx} = -y^{-2}\frac{dy}{dex}$

$\displaystyle \Rightarrow -\frac{dz}{dx}+z=1$

$\displaystyle \Rightarrow \frac{dz}{dx}+(-1)z=-1$

$\displaystyle I = e^{\int{-1 dx}} = e^{-x}$

$\displaystyle e^{-x}z = \int{-1 dx}$

$\displaystyle \Rightarrow e^{-x}z = -x + C$

$\displaystyle \Rightarrow z = e^{x}(-x + C)$

$\displaystyle \Rightarrow y^{-1} = e^{x}(-x + C)$

$\displaystyle \Rightarrow \frac{1}{e^{x}(-x + C)}$

The correct answer given by the professor is:
$\displaystyle \frac{1}{1+Cx}$

The standard approach for solving the DE...

$\displaystyle y^{'} = -\frac{2}{x}\ y + y^{2}$ (1)

... has as first step the division of both terms of (1) by $\displaystyle y^{2}$ obtaining...

$\displaystyle \frac{y^{'}}{y^{2}} = -\frac{2}{x\ y}+1$ (2)

... and then the substution...

$\displaystyle z= \frac{1}{y} \implies z^{'}= -\frac{y^{'}}{y^{2}}$ (3)

... so that the new DE in z is...

$\displaystyle z^{'}= \frac{2}{x}\ z -1$ (4)

... which is linear and easy enough to solve...

... the minor problem is that (4) seems not the same You obtained that is...

$\displaystyle z^{'}= z-1$ (5)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 5th 2011, 08:38 AM
Mike9182
Re: Bernoulli's Equation
I fixed that step, although I am still having trouble finding the correct answer.

Original DE:
$\displaystyle \frac{dy}{dx} + \frac{2}{x}y = y^2$

$\displaystyle \Rightarrow (1) \frac{1}{y^2}\frac{dy}{dx} + \frac{2}{xy^2}y = 1$

$\displaystyle \Rightarrow (2) \frac{1}{y^2}\frac{dy}{dx}+\frac{2}{x}y^{-1} = 1$

$\displaystyle (3) z = \frac{1}{y}$
$\displaystyle (4) z' = \frac{-1}{y^2}y'$

$\displaystyle \Rightarrow (5) -z' + \frac{2}{x}z = 1$

$\displaystyle \Rightarrow (6) z' + (-\frac{2}{3})z = -1$

$\displaystyle (7) I = e^{\int{-\frac{2}{3}dx}} = e^{-\frac{2}{3}x}$

$\displaystyle (8) e^{-\frac{2}{3}}z = \int{e^{-\frac{2}{3}x}(-1)dx} = \frac{3}{2}e^{-\frac{2}{3}x}$

$\displaystyle \Rightarrow (9) z = \frac{3}{2}$

$\displaystyle \Rightarrow (10) \frac{1}{y} = \frac{3}{2}$

$\displaystyle \Rightarrow (11) y = \frac{2}{3}$

I incorrectly typed the answer the professor gave me in my first post. The answer he gave is:
$\displaystyle \frac{1}{x(1 + Cx)}$
• Sep 6th 2011, 04:44 AM
Jester
Re: Bernoulli's Equation
From equation (5) to (6). Why did you goe from 2/x to 2/3?
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