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