MA209 Mathematics Homework Assignment

Running head: MA209 HOMEWORK 2
MA209 HOMEWORK 2
Name of the Student
Name of the University
Author Note

MA209 HOMEWORK 21
Question 1:
a) dy/dt = (y-2)(y-5), y(0) = 0
=> dy/((y-2)(y-5)) = dt
=> (1/3)((y-2)-(y-5))dy/((y-2)(y-5)) = dt
=> 1
3 ∫ ( ( y−2 ) − ( y−5 ) ) dy
( y−2 ) ( y−5 ) =∫dt
=> 1
3 ∫ 1
y−5 d ( y−5 )− 1
3 ∫ d ( y −2 )
y−2 =∫ dt
=> (1/3)ln(y-5) – (1/3)ln(y-2) = t + ln(c)
=> (1/3)ln((y-5)/c(y-2)) = t
=> y −5
c ( y −2 ) =exp ( 3 t )
Now, putting initial value y(0) = 0
 -5/(c*(-2)) = 1
 c = 5/2
Hence, the solution is
y−5
( 5
2 ) ( y−2 )
=exp ( 3 t )
 y-5 = ( 5 y
2 )exp (3 t )−5 exp (3 t)
 y(1-(5/2)exp ( 3 t )) = 5 - 5 exp ( 3t )
 y = 10 ( 1−exp ( 3 t ) )
2−5 exp ( 3 t ) = 10 ( exp (−3 t )−1 )
2exp (−3 t)−5 = 10 ( 1−exp (−3 t ) )
5−2 exp (−3 t)

MA209 HOMEWORK 22
Hence, option (iii) is correct.
b)
The generalized Heun’s algorithm is implemented in MATLAB as given below.
MATLAB code:
function [t,y] = q1_1805342(T,N)
func = @(t,y) (y-2)*(y-5);
t = linspace(0,T,N+1);
h = t(2) - t(1);
y(1) = 0;
for i=1:length(t)-1
y(i+1) = y(i) + (h/2)*(func(t(i),y(i)) + func(t(i+1),y(i) + h*func(t(i),y(i)))); % Modified
Euler's method or Heun's Algorithm with alpha = 1/2
end
end
c)
Now, for finding the N a MATLAB function is written where error tolerance is taken as input
and the minimum N for reaching below that tolerance and the step size is displayed.
i)
[N,h] = q1c(1e-2)
N =

MA209 HOMEWORK 23
188
h =
0.0535
Hence, the value of N = 188.
ii)
[N,h] = q1c(1e-4)
N =
1734
h =
0.0058
iii)
Now, when h = 0.0535 then err <~ 0.02
h^3 = 1.5313e-04
err = K*1.5313e-4
 K = 130.608
when h = 0.0058 then err <~ 1e-4.
h^3 = 1.9511e-07
err = K*1.9511e-07
 K = 512.5314
Hence, as K < 1000 hence, modifier Euler is O(h^3) accurate.