Solving Bernoulli Differential Equation.

• January 1st 2012, 01:32 AM
amro05
Solving Bernoulli Differential Equation.
• January 1st 2012, 01:40 AM
Prove It
Re: NEED TO SOLVE >>> bernoulli equation
Quote:
Start by making the substitution \displaystyle \begin{align*} v = y^{1-n} \end{align*}.
• January 1st 2012, 01:44 AM
amro05
Re: NEED TO SOLVE >>> bernoulli equation
Quote:

Originally Posted by prove it
start by making the substitution \displaystyle \begin{align*} v = y^{1-n} \end{align*}.

yes sir i did it

but the problem is the absulet number (( 2 )) << without it , it became easy
• January 1st 2012, 02:17 AM
chisigma
Re: NEED TO SOLVE >>> bernoulli equation
Quote:
Applying the 'standard procedure' the substitution $w=\frac{1}{y}$ leads to the DE...

$w^{'}= - 1+2 w x -2 w^{2}$ (1)

... which is known as 'Riccati Equation'. The solution of this type of equation is usually an hard task but, may be, not impossible in this case. The question requires further investigation...

Kind regards

$\chi$ $\sigma$
• January 1st 2012, 05:54 AM
amro05
Re: NEED TO SOLVE >>> bernoulli equation
Quote:

Originally Posted by chisigma
Applying the 'standard procedure' the substitution $w=\frac{1}{y}$ leads to the DE...

$w^{'}= - 1+2 w x -2 w^{2}$ (1)

... which is known as 'Riccati Equation'. The solution of this type of equation is usually an hard task but, may be, not impossible in this case. The question requires further investigation...

Kind regards

$\chi$ $\sigma$

thank you sir for your help

I begin with
http://photoserver.ws/images/kaqR4f007165bc1cc.jpg

and I start with y1=2 and with y1=2x >>> but both gives more complicate Eq.

thank you sir for your help
• January 1st 2012, 07:33 AM
chisigma
Re: NEED TO SOLVE >>> bernoulli equation
Quote:
The original DE...

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

... is itself a 'Riccati Equation' so that a substitution of the type $w=\frac{1}{y}$ is useless. More promising is the substiution $w=y-x$ after that the (1) becomes...

$w^{'}=1-x^{2} + w^{2}$ (2)

The (2) is also a 'Riccati Equation' but, very important detail, we can verify that $w_{1}=x$ is solution of (2) and that allows You to find the general solution as follows. Setting $w=w_{1}+u=x+u$ the (2) becomes...

$u^{'}=2 u x + u^{2}$ (3)

... which is a 'Bernoulli Equation', the solution of which is left to You...

Kind regards

$\chi$ $\sigma$
• January 1st 2012, 10:46 AM
amro05
Re: NEED TO SOLVE >>> bernoulli equation
Quote:

Originally Posted by chisigma
Kind regards

$\chi$ $\sigma$

(Clapping)
Thank you very very mach .
all the best

(Hi)