Course:MATH102/Question Challenge/2001 December Q7

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Question

Solutions to the differential equation

$\displaystyle {\frac {dy}{dt}}=a-by$

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

Solution curves

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

[show]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 $y$?

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

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

[show]Hint 4:
blah

Solutions

[show]Solution

Part A) The slopes $m_{0},m_{1},m_{2}$ have to match with the value of $dy/dt$ when $y=0,1,2$ respectively, which provides a number of equations. We know that: $m_{1}=a-b(1)=a-b$ $m_{1}=a-b(2)=a-2b$ We also know: $m_{0}=a-b(0)=a$ we are given that $m_{0}=5m_{2}$ and $m_{1}=3$ Therefore by extension: $a=5(a-2b)$ and $a-b=3$ The first of these can also be written as $4a=10b$ after simplification. We now have two equations in two unknowns, $a,b$. $4a=10b$ $a-b=3$ 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 ${\frac {dy}{dx}}=0$: ${\frac {dy}{dx}}=a-by$ $0=(5)-(2)y$ Solving for $y$: $2y=5$ $y={\frac {5}{2}}$ A phase graph would show that for values of $y<{\frac {5}{2}}$ the rate of change $dy/dt>0$. These values will move to the right towards $y={\frac {5}{2}}$. For values of $y>{\frac {5}{2}}$ the rate of change $dy/dt<0$. 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 ${\frac {dy}{dt}}=a-by$ are given by: $y(t)={\frac {a}{b}}+(y_{0}-{\frac {a}{b}})e^{-bt}$ 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