I have the following differential equation:
$$ y'(t) + ay(t) = de^{-bt} $$
with $a,b,d$ constants and $a,b>0$ and I need to calculate:
$$\lim_{t\to \infty} y(t)$$ (hint consider a is not equal to b, and a equals b separately, a hint the book gives)
ATTEMPT:
So far I have found the general solution:
$$ y = de^{-at} \left(\frac{e^{t(a-b)}}{a-b} + C \right )$$
I am confused about how to take the limit of the problem? Do I plug in infinity where all the $t$'s are and solve for $C$ (constant from integration)? I am not sure where to go..
Thanks for the help!
As Andrei wrote in the comment rewrite $y(t)$ as:
For $(a \ne b)$: $$y(t) = de^{-at} \left(\frac{e^{t(a-b)}}{a-b} + C \right )$$ $$y(t) = \frac d{e^{bt}(a-b)} + \frac C {e^{at}}$$ Plug infinity in the expression to calculate the limit. ( don't forget that $a,b > 0)$. The result should be zero. Since both $a,b >0$.
For $a=b$: $$y′(t)+ay(t)=de^{-at}$$ $$(y(t)e^{at})'=d$$ $$y(t)e^{at}=dt+C$$ $$y(t)=\frac {dt+C}{e^{at}}$$ You have a polynomial at the numerator and an exponential at the denominator $(a>0)$ so the limit is zero at infinity. $$\lim\limits_{t\to\infty}\{y(t)\}=0$$