# Bernoulli's Equation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by Mike9182
What did I do wrong here?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The standard approach for solving the DE...

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

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

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

... and then the substution...

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$
• September 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:
$\frac{dy}{dx} + \frac{2}{x}y = y^2$

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

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

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

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

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

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

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

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

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

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

I incorrectly typed the answer the professor gave me in my first post. The answer he gave is:
$\frac{1}{x(1 + Cx)}$
• September 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?