a peculiar ODE

**mmzaj** · December 31st 2009, 06:02 PM

what is the solution for y in this peculiar ODE ?

$A\left(y,x\right)=\frac{dy}{dx}+B(x)(1-y)$

with initial conditions :

$\frac{dy}{dx}= 0 \ldots,y=0$

$\frac{dy}{dx}=\delta(x-x_{0})\ldots,y=1$

moreover

$\int^{\infty}_{-\infty}Adx=\int^{\infty}_{-\infty}Bdx=1$

**chisigma** · January 1st 2010, 06:33 AM

The condition...

$\int_{-\infty}^{+\infty} A(x,y)\cdot dx=1$ (1)

... means that $A(*,*)$ is a function of the only $x$ so that the DE is...

$\frac{dy}{dx} = \alpha(x)\cdot y + \beta (x)$ (2)

... where ...

$\alpha(x)= B(x)$

$\beta(x)= A(x) - B(x)$ (3)

The (2) is a linear first order DE and its general solution is...

$y(x)= e^{\int \alpha(x)\cdot dx} \{\int \beta(x)\cdot e^{-\int \alpha(x)\cdot dx}\cdot dx + c\}$ (4)

The constant c is derived [when possible...] from the 'initial conditions' ...

Kind regards

$\chi$ $\sigma$

**mmzaj** · January 1st 2010, 06:34 PM

Originally Posted by chisigma

The condition...

$\int_{-\infty}^{+\infty} A(x,y)\cdot dx=1$ (1)

... means that $A(*,*)$ is a function of the only $x$ so that the DE is...

$\frac{dy}{dx} = \alpha(x)\cdot y + \beta (x)$ (2)

... where ...

$\alpha(x)= B(x)$

$\beta(x)= A(x) - B(x)$ (3)

The (2) is a linear first order DE and its general solution is...

$y(x)= e^{\int \alpha(x)\cdot dx} \{\int \beta(x)\cdot e^{-\int \alpha(x)\cdot dx}\cdot dx + c\}$ (4)

The constant c is derived [when possible...] from the 'initial conditions' ...

Kind regards

$\chi$ $\sigma$

thanks , that was really helpful . but i stated the simplest form the problem in hand , where A , B , and y are functions of the single variable x . but the real case requires that the latter functions could be functions of an arbitrary number of independent variables $x_{1},x_{2}..x_{n}$ and the equation becomes :

$A(x_{1},x_{2}..x_{n})=\frac{dy}{d\Omega}+B(x_{1},x _{2}..x_{n})(1-y)$
where $d\Omega=\prod^{n}_{i=1} dx_{i}$
is a unit volume (so to speak !!)

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