Class 12 Differential Equations · Case Study4 Marksseparable_differential_equations
Class 12 Maths · Differential Equations · Case Study #118
A district has 50 thousand children. In an idealized model, the rate of administering drops is proportional to the number not yet covered. Let x be time in weeks and y the number covered in thousands: y′ = k(50 − y). Initially y(0) = 0, and y(2) = 25. Hence k = (ln 2)/2 per week. Answer the following mathematical modelling questions.
Part (1) 1 Marks
Find the general solution of y′ = k(50 − y).
Answer: y = 50 − Ae^(−kx), A ∈ ℝ.
- Set z = 50 − y, so z′ = −kz. The linear equation gives z = Ae^(−kx), including A = 0.
- Thus y = 50 − Ae^(−kx); the application uses k > 0 and 0 ≤ y ≤ 50.
Part (2) 1 Marks
Using −ln|50 − y| = kx + C and y(0) = 0, find C.
Answer: C = −ln 50.
- At x = 0, y = 0, substitute into −ln|50 − y| = kx + C.
- C = −ln 50, independently of the numerical value of k.
Part (3) 2 Marks
Find the particular model and use it to estimate the number covered after three weeks.
Answer: y(x) = 50(1 − 2^(−x/2)) thousand; y(3) ≈ 32.322 thousand, about 32,322 children.
- From y = 50 − Ae^(−kx), y(0) = 0 gives A = 50.
- The condition y(2) = 25 gives e^(−2k) = 1/2, so k = (ln 2)/2 and y(x) = 50(1 − 2^(−x/2)).
- At x = 3, y(3) = 50(1 − 1/(2√2)) ≈ 32.322 thousand, about 32,322 children.