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- Jan 1st 2012, 01:32 AMamro05Solving Bernoulli Differential Equation.
- Jan 1st 2012, 01:40 AMProve ItRe: NEED TO SOLVE >>> bernoulli equation
- Jan 1st 2012, 01:44 AMamro05Re: NEED TO SOLVE >>> bernoulli equation
- Jan 1st 2012, 02:17 AMchisigmaRe: NEED TO SOLVE >>> bernoulli equation
Applying the 'standard procedure' the substitution $\displaystyle w=\frac{1}{y}$ leads to the DE...

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

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 1st 2012, 05:54 AMamro05Re: NEED TO SOLVE >>> bernoulli equation
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 - Jan 1st 2012, 07:33 AMchisigmaRe: NEED TO SOLVE >>> bernoulli equation
The original DE...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 1st 2012, 10:46 AMamro05Re: NEED TO SOLVE >>> bernoulli equation