Answer
$48$ $g/L$
Work Step by Step
$\frac{dy}{dt}$ = salt rate in - salt rate out
$\frac{dy}{dt}$ = $(80)(50)-(120)(\frac{y}{500-40t})$
$\frac{dy}{dt}$ = $4000-\frac{12y}{50-4t}$
$\frac{dy}{dt}+\frac{6y}{25-2t}$ = $4000$
$A(t)$ = $\frac{6}{25-2t}$
$B(t)$ = $4000$
by theorem 1, $α(x)$ is defined by
$α(t)$ = $e^{\int{\frac{6}{25-2t}}dt}$ = $e^{-3\ln{(25-2t)}}$ = $(25-2t)^{-3}$
$[(25-2t)^{-3}y]'$ = $4000(25-2t)^{-3}$
$[(25-2t)^{-3}y]$ = $1000(25-2t)^{-2}+C$
$y(t)$ = $25000-2000t+C(25-2t)^{3}$
$y(0)$ = $10$
so
$10$ = $25000+C(50)^{3}$
$C$ = $-1.599$
$y(t)$ = $25000-2000t-1.599(25-2t)^{3}$
the amount of salt at $t$ = $10$
$y(10)$ = $4800.08$ $g$
so the concentration of salt is
$\frac{4800.08g}{1000L}$ = $48$ $g/L$
