
Yefim S. answered 02/18/21
Math Tutor with Experience
Let q(t)grams is mass of salt in tank in moment t.
Then change of mass of salt dq/dt = 0.5c - 1q/(10 - 0.5t);
q' + (1/(10 - 0.5t))q = 0.5c, c g/L is unknown concentration.
We get linear ODE. Integrated factor M = e∫1/(10 - 0.5t)dt = e- 2ln(10 - 5t) = (10 - 0.5t)-2.
Multiplying by m our ODE: (10 - 0.5t)-2q' + (10 - 0.5t)-3q = 0.5c(10 - 0.5t)-2;
[(10 - 0.5t)-2q]' = 0.5c(10 - 0.5t)-2; (10 - 0.5t)-2q = ∫0.5c(10 - 0.5t)-2dt;
(10 - 0.5t)-2q = c(10 - 0.5t)-1 + N; q = c(10 - 0.5t) + N(10 - 0.5t)2; at t = 0 q= 0; 0 = 10c + 100N, so N = - 0.1c
At last q = c(10 - 0.5t) - 0.1c(10 - 0.5t)2; If t = 20 min q = 10 g;
But by this data c is undefined.