Course:MATH102/Question Challenge/2001 December Q7

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Question

Solutions to the differential equation

$\frac{d y}{d t} = a - b y$

starting at three different initial values are shown in the graph. Also shown are the tangent lines to these curves at $t = 0$

You are given the following information about the slopes of these tangent lines:

(i) The slope $m_{0}$ is five times the slope $m_{2}$ .

(ii) The slope $m_{1}$ is 3.

Use this information to answer the following questions:

(a) Determine the values of the constants a and b. (Justify your answer).

(b) Determine the value that $y (t)$ will approach after a long time on any of these curves.

(c) Find the value of $y (t)$ at time $t = 0.2$ given that $y (0) = 1$ . Justify how your answer was obtained.

Hints

Hint 1:
What is the differential equation saying about the slope of a tangent line to each solution curve? How is that slope related to the current value of $y$ ?

Hint 2:
Look at the values of $y$ at t=0 on each of the curves and relate $m_{i} = \frac{d y}{d t}$ to $a, b$ and $y$ .

Hint 3:
You should get some equations for $a, b$ . Solve those equations to find the values of these constants

Hint 4:
blah

Solutions

Solution
Part A) The slopes $m_{0}, m_{1}, m_{2}$ have to match with the value of $d y / d t$ when $y = 0, 1, 2$ respectively, which provides a number of equations. We know that: $m_{1} = a - b (1) = a - b$ $m_{1} = a - b (2) = a - 2 b$ We also know: $m_{0} = a - b (0) = a$ we are given that $m_{0} = 5 m_{2}$ and $m_{1} = 3$ Therefore by extension: $a = 5 (a - 2 b)$ and $a - b = 3$ The first of these can also be written as $4 a = 10 b$ after simplification. We now have two equations in two unknowns, $a, b$ . $4 a = 10 b$ $a - b = 3$ We can solve these to obtain $a = 5, b = 2$ Part B) To find the value $y (t)$ will approach after a very long time on any of these curves we must find a stable steady state. First we will set $\frac{d y}{d x} = 0$ : $\frac{d y}{d x} = a - b y$ $0 = (5) - (2) y$ Solving for $y$ : $2 y = 5$ $y = \frac{5}{2}$ A phase graph would show that for values of $y < \frac{5}{2}$ the rate of change $d y / d t > 0$ . These values will move to the right towards $y = \frac{5}{2}$ . For values of $y > \frac{5}{2}$ the rate of change $d y / d t < 0$ . These values will moves to the left towards $y = \frac{5}{2}$ . This describes a stable steady state so these curves will approach this value of $y (t) = \frac{5}{2}$ over time. Part C) Solution to differential equation of the form $\frac{d y}{d t} = a - b y$ are given by: $y (t) = \frac{a}{b} + (y_{0} - \frac{a}{b}) e^{- b t}$ Solving for $y (0.2)$ : $y (0.2) = \frac{5}{2} + ((1) - \frac{5}{2}) e^{- 2 (0.2)}$ $y (0.2) = \frac{5}{2} - \frac{3}{2} e^{- 0.4}$ Comment: A solution using Euler's method was also acceptable and would have resulted in a solution of 1.54