Solving for y in an ODE with Initial Conditions

[mmzaj]

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}=\left0 \ldots , y=0

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

moreover

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

 

[berkeman]

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}=\left0 \ldots , y=0

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

moreover

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

What is the context of your question? Is this schoolwork?

 

[HallsofIvy]

"A(x,y)" implies that A is a function of the independent variables x and y but then dy/dx makes no sense.

 

[mmzaj]

What is the context of your question? Is this schoolwork?

yes and no , it's not a homework , it's for a term paper . the functions A&B are PDFs of some kind , and y is a cumulative distribution function .

 

[mmzaj]

ok , here is a trial :

\frac{dy}{dx} = \alpha(x)\cdot y + \beta (x) (1)

where ...

\alpha(x)= B(x)

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

(1) 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)

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

 

[mmzaj]

"A(x,y)" implies that A is a function of the independent variables x and y but then dy/dx makes no sense.

you can think of \frac{dy}{dx} as an implicit derivative rather than explicit