Answer
a) $y(t)$ = $\frac{200000t+8000t^{2}+500}{50+4t}$
b) $37.505$ $g/L$
Work Step by Step
$\frac{dy}{dt}$ = salt rate in - salt rate out
$\frac{dy}{dt}$ = $(80)(50)-(40)(\frac{y}{500+40t})$
$\frac{dy}{dt}$ = $4000-40\frac{y}{500+40t}$
$\frac{dy}{dt}+40\frac{y}{500+40t}$ = $4000$
$A(x)$ = $\frac{4}{50+4t}$
$B(x)$ = $4000$
by theorem 1, $α(x)$ is defined by
$α(x)$ = $e^{\int{\frac{4}{50+4t}}dt}$ = $e^{\ln{(50+4t)}}$ = $50+4t$
$[(50+4t)y]'$ = $4000(50+4t)$
$[(50+4t)y]$ = $200000t+8000t^{2}+C$
$y(t)$ = $\frac{200000t+8000t^{2}+C}{50+4t}$
$y(0)$ = $10$
so
$10$ = $\frac{C}{50}$
$C$ = $500$
$y(t)$ = $\frac{200000t+8000t^{2}+500}{50+4t}$
b)
The tank overflows when t = $\frac{25}{2}$ = $12:5$. The amount of salt in the tank at that time is
$y(12.5)$ = $37505$ g
so the concentration of salt is
$\frac{37505g}{1000L}$ = $37.505$ $g/L$
