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!
$p'(t)+ yp(t)= xe^{-xt}$ 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 $-Axe^{-xt}+ Aye^{-xt}= A(y- x)e^{-xt}= xe^{-xt}$. $A= \frac{x}{y- x}$.
$p(t)= Ce^{-yt}+ \frac{x}{y- x}e^{-xt}$.