1. exponential integrals

q(t)=exp(-xt)

p'(t)=xq(t)-yp(t)

In need to find p(t) .

I know p(t)=(x/y-x)(exp(-xt)-exp(-yt)) but don't know how to get to it.

I would appreciate your help, thank you!

2. Re: exponential integrals

use $e^{y t}$ as an integrating factor

$e^{y t} p^\prime + y e^{y t}p = x e^{y t}e^{-x t}$

$(e^{y t}p)^\prime = x e^{(y-x)t}$

etc.

I leave you to complete it, it should be straightforward from here.

3. Re: exponential integrals

$p'(t)+ yp(t)= xe^{-xt}$ is a linear, first order, non-homogenous equation with constant coefficients (since p is a function of t only, we can treat x and y as constants). Its characteristic equation is $r+ y= 0$ so r= -y. The general solution to the associated homogeneous equation is $p(t)= Ce^{-yt}$. We look for a specific solution to the entire equation of the form $p(t)= Ae^{-xt}$. $y(t)'= -Axe^{-xt}$ so the equation becomes $-Axe^{-xt}+ Aye^{-xt}= A(y- x)e^{-xt}= xe^{-xt}$. $A= \frac{x}{y- x}$.

$p(t)= Ce^{-yt}+ \frac{x}{y- x}e^{-xt}$.