Solving the General Bernoulli / Riccati Equation

[crevoise]

Hello,

I have an equation of the following form: y'(t) + Ay(t) + By(t)^m + c = 0

1/ With c=0, the equation is of the Bernoulli form, and might be integrated.
2/ With m = 2, it is a Riccati equation, which might be turn back to a Bernoulli one and then integrated

3/ My question is about the general case, with c#0 and m a real (not only a natural). Does it exist some way to solve this general equation? Maybe there is a method to turn it back to Riccati or Benoulli shape?

Thanks a lot for your help

/crevoise

 

[chisigma]

Hello,

I have an equation of the following form: y'(t) + Ay(t) + By(t)^m + c = 0

1/ With c=0, the equation is of the Bernoulli form, and might be integrated.
2/ With m = 2, it is a Riccati equation, which might be turn back to a Bernoulli one and then integrated

3/ My question is about the general case, with c#0 and m a real (not only a natural). Does it exist some way to solve this general equation? Maybe there is a method to turn it back to Riccati or Benoulli shape?

Thanks a lot for your help

/crevoise

Let write the DE as...

$\displaystyle \frac{y^{\ '}+c}{y^{m}} + \frac{a}{y^{m-1}}=-b$ (1)

First we set...

$\displaystyle u=y+c\ x \implies u^{\ '}= y^{\ '} + c\ ,\ y^{m}= (u-c\ x)^{m} $ (2)

... so that (1) becomes...

$\displaystyle \frac{u^{\ '}} {(u-c\ x)^{m}} + \frac{a} {(u-c\ x)^{m-1}}= -b$ (3)

Now we set...

$\displaystyle v=\frac{1}{(u-cx)^{m-1}} \implies v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}\ u^{\ '}$ (4)

... and if $m \ne 1$ the (3) becomes...

$\displaystyle \frac{v^{\ '}}{1-m} + a\ v = -b$ (5)

... which is linear...

Kind regards

$\chi$ $\sigma$

 

[crevoise]

Let write the DE as...

$\displaystyle \frac{y^{\ '}+c}{y^{m}} + \frac{a}{y^{m-1}}=-b$ (1)

First we set...

$\displaystyle u=y+c\ x \implies u^{\ '}= y^{\ '} + c\ ,\ y^{m}= (u-c\ x)^{m} $ (2)

... so that (1) becomes...

$\displaystyle \frac{u^{\ '}} {(u-c\ x)^{m}} + \frac{a} {(u-c\ x)^{m-1}}= -b$ (3)

Now we set...

$\displaystyle v=\frac{1}{(u-cx)^{m-1}} \implies v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}\ u^{\ '}$ (4)

... and if $m \ne 1$ the (3) becomes...

$\displaystyle \frac{v^{\ '}}{1-m} + a\ v = -b$ (5)

... which is linear...

Kind regards

$\chi$ $\sigma$

Hello,

Thanks a lot for your answer.
I think yet there is a small mistake on it, on the second change:

$\displaystyle v=\frac{1}{(u-cx)^{m-1}} \implies v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}\ u^{\ '}$ (4)

I think it implies

$\displaystyle v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}(\ u^{\ '}-c)$

isn't it?

 

[Mathwebster]

Let write the DE as...

$\displaystyle \frac{y^{\ '}+c}{y^{m}} + \frac{a}{y^{m-1}}=-b$ (1)

First we set...

$\displaystyle u=y+c\ x \implies u^{\ '}= y^{\ '} + c\ ,\ y^{m}= (u-c\ x)^{m} $ (2)

... so that (1) becomes...

$\displaystyle \frac{u^{\ '}} {(u-c\ x)^{m}} + \frac{a} {(u-c\ x)^{m-1}}= -b$ (3)

Now we set...

$\displaystyle v=\frac{1}{(u-cx)^{m-1}} \implies v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}\ u^{\ '}$ (4)

... and if $m \ne 1$ the (3) becomes...

$\displaystyle \frac{v^{\ '}}{1-m} + a\ v = -b$ (5)

... which is linear...

Kind regards

$\chi$ $\sigma$

You might want to check step (4).

 

[Mathwebster]

To the OP - Are A, B and C constant.

 

[crevoise]

Yes, A, B and C are constant for the moment.

 

[Mathwebster]

Then your ODE is separable. A much different animal if $A, B$ and $C$ are functions of $x$.

 

[chisigma]

Let write the DE as...

$\displaystyle \frac{y^{\ '}+c}{y^{m}} + \frac{a}{y^{m-1}}=-b$ (1)

First we set...

$\displaystyle u=y+c\ x \implies u^{\ '}= y^{\ '} + c\ ,\ y^{m}= (u-c\ x)^{m} $ (2)

... so that (1) becomes...

$\displaystyle \frac{u^{\ '}} {(u-c\ x)^{m}} + \frac{a} {(u-c\ x)^{m-1}}= -b$ (3)

Proceeding in correct way from (3) now we set...

$\displaystyle v=\frac{1}{(u-cx)^{m-1}} \implies v^{\ '}= \frac{1-m}{(u-c\ x)^{m}}\ (u^{\ '}-c)$ (4)

... so that if $m \ne 1$ the (3) becomes in some steps...

$\displaystyle v^{\ '}= (m-1)\ (a\ v+ c\ v^{\frac{m}{m-1}}+ b)$ (5)

... where the variables are separated...

Kind regards

$\chi$ $\sigma$