Answer
$50$ $g/L.$
Work Step by Step
$\frac{dy}{dt}$ = salt rate in - salt rate out
$\frac{dy}{dt}$ = $(80)(50)-(80)(\frac{y}{500})$
$\frac{dy}{dt}$ = $4000-\frac{8y}{50}$
$\frac{dy}{dt}+\frac{8y}{50}$ = $4000$
$A(x)$ = $\frac{8}{50}$
$B(x)$ = $4000$
by theorem 1, $α(x)$ is defined by
$α(x)$ = $e^{\int{\frac{8}{50}}dt}$ = $e^{0.16t}$ = $50+4t$
$[(e^{0.16t})y]'$ = $4000(e^{0.16t})$
$[(e^{0.16t}y]$ = $25000e^{0.16t}+C$
$y$ = $25000+\frac{C}{e^{0.16t}}$
As $t \to \infty$, the exponential term tends to zero, so that the amount of salt tends to $25000g$, or $50$ $g/L.$
