a. Salt flows into tank A at a rate of
(y(t)/500 lb/gal) * (50 gal/min) = y(t)/10 lb/gal
and out at a rate of
(x(t)/500 lb/gal) * (50 gal/min) = x(t)/10 lb/gal
so the net rate of change of the amount of salt in tank A is
[tex]x'(t)=\dfrac{y(t)}{10}-\dfrac{x(t)}{10}[/tex]
Similarly, you would find
[tex]y'(t)=\dfrac{x(t)}{10}-\dfrac{y(t)}{10}[/tex]
b. We have
[tex]x'=-\dfrac x{10}+\dfrac y{10}\implies x''=-\dfrac{x'}{10}+\dfrac{y'}{10}[/tex]
Notice that x' = -y', so
[tex]x''=-{2x'}{10}\implies 5x''+x'=0[/tex]
Solve for x: the characteristic equation
[tex]5r^2+r=r(5r+1)=0[/tex]
has roots [tex]r=0[/tex] and [tex]r=-\frac15[/tex], so
[tex]x(t)=C_1+C_2e^{-t/5}[/tex]
Again using the fact that y' = -x', we then find
[tex]x'(t)=-\dfrac{C_2}5e^{-t/5}\implies y'(t)=\dfrac{C_2}5e^{-t/5}\implies y(t)=C_1-C_2e^{-t/5}[/tex]
Given that x(0) = 0 and y(0) = 10, we find
[tex]\begin{cases}0=C_1+C_2\\10=C_1-C_2\end{cases}\implies C_1=5,C_2=-5[/tex]
