Definite Integral of p(x,y) with Bounds x>y>0 and dp(x,y)/dx = (e^-x^2)

[IsomaDan]

Homework Statement

The function, p(x;y), of two variables is defined for x>y>0, and satisfies

We furthermore know that dp(x,y)/dx = (e^-x^2)

and that p(y; y) = 0

Homework Equations

I now need to write p(x,y) as a definite integral of the form int (f(t)dt, with lower bound t=H and upper bound x.

The Attempt at a Solution

I suppose I need the info p(y; y) = 0 to get the bounds, but not quite sure how. I sat earlier an evaluated the integral from 0 to inf and from -inf to inf, however I do not see how that could come in handy. I am thinking of treating the function as one of x alone as a start, but do not know how to proceed from there?
Anyone can give me a hint :-) Th

 

[SammyS]

Homework Statement

The function, p(x;y), of two variables is defined for x>y>0, and satisfies   ?  

We furthermore know that dp(x,y)/dx = (e^-x^2)

and that p(y; y) = 0

Homework Equations

I now need to write p(x,y) as a definite integral of the form int (f(t)dt, with lower bound t=H and upper bound x.

The Attempt at a Solution

I suppose I need the info p(y; y) = 0 to get the bounds, but not quite sure how. I sat earlier an evaluated the integral from 0 to inf and from -inf to inf, however I do not see how that could come in handy. I am thinking of treating the function as one of x alone as a start, but do not know how to proceed from there?
Anyone can give me a hint :-) Th

Do you really mean x>y>0, or di you mean x>0 and y>0 ?

Does p(x,y) satisfy something in addition to what's stated ?

Check for any other typos you may have, so that your post is more readable.

 

[IsomaDan]

I meant satisfies "x ≥ y ≥ 0". Don't know where that went.

Sorry for that. All the best

Jonas

 

[haruspex]

I meant satisfies "x ≥ y ≥ 0".

Still doesn't make sense. That's a change to the specified domain of p, not a property of p.