MA209 Mathematics Homework Assignment

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

1MA209 HOMEWORK 2
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)

2MA209 HOMEWORK 2
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 =

3MA209 HOMEWORK 2
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.

4MA209 HOMEWORK 2
Hence,
Err <~ O(h^3)
d) The limiting value of y(t) as t approaches ∞ is found from a sample run.
Sample run:
[t,y] = q1_1805342(10,100)
Plot:
0 1 2 3 4 5 6 7 8 9 10
t
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
y(t)
Hence, lim
t → ∞
y (t)=2
Question 2:
a)
MATLAB function:

5MA209 HOMEWORK 2
function [x,y] = q2_1805342(T,N)
x(1) = 1; xmax = 1000; betax = 1;
y(1) = 7; ymax = 1000; betay = 1;
% function of differential equation
fx = @(x,y,t) betax*x*(1-x/xmax - y/ymax);
fy = @(x,y,t) betay*y*(1-x/xmax - y/ymax);
t = linspace(0,T,N+1);
h = t(2) - t(1);
for i=1:length(t)-1
% Euler's formula
x(i+1) = x(i) + h*fx(x(i),y(i),t(i));
y(i+1) = y(i) + h*fy(x(i),y(i),t(i));
end
plot(t,x,t,y)
title('Solution by Euler Method')
legend('x(t)','y(t)')
xlabel('time t in secs')
ylabel('Population')
end

6MA209 HOMEWORK 2
Sample run:
q2_1805342(20,500)
Plot:
0 2 4 6 8 10 12 14 16 18 20
time t in secs
0
100
200
300
400
500
600
700
800
900
Population
Solution by Euler Method
x(t)
y(t)
b)
i) The reasonable values of T is such that the start to final behaviour of the solution can be
observed in detail. Thus T = 50 is chosen.
ii) Now, for maintaining accuracy the curves of x(t) and y(t) should be smooth enough and
hence N is chosen to be N = 500.

7MA209 HOMEWORK 2
c) x¿ = lim
t → ∞
x ( t )=¿125
y¿=lim
t → ∞
y ( t ) = 875
d) The limiting values x¿∧¿ y¿ depends on the initial values x0 and y0 in the following way
when xmax = ymax = 1000.
x∗¿ 1000∗( x 0
x 0+ y 0 )
y∗¿1000∗( y 0
x 0+ y 0 )
Question 3:
The direct polynomial interpolation method is applied fit the given data.
MATLAB code:
function [coeff,xdata,ydata] = q3_1805342(x,y)
mat = ones(length(x),length(x));
for i=2:length(x)
mat(:,i) = x.^(i-1)';
end
coeff = mat^(-1)*y'; % finding coeffcients of polymial function
xdata = min(x):0.1:max(x);
ydata = zeros(1,length(xdata));
for i=1:length(coeff)

8MA209 HOMEWORK 2
ydata = ydata + coeff(i).*xdata.^(i-1); % evaluating polymial at x points
end
plot(x,y,'ro',xdata,ydata,'b-')
grid on
xlabel('x values')
ylabel('y values')
legend('given data points','fitted polynomial')
end
Output:
>> x = [-3,-2,-1,1,2,3];
>> y = [546,240,50,-150,-304,-390];
>> q3_1805342(x,y)
ans =
-48.0000
-84.0000
-4.0000
-17.0000
2.0000
1.0000
Plot:

9MA209 HOMEWORK 2
-3 -2 -1 0 1 2 3
x values
-400
-300
-200
-100
0
100
200
300
400
500
600
y values
given data points
fitted polynomial
b) From the above output the calculated coefficients for the polynomial function is given by
p(x) = -48 -84x – 4x^2 -17x^3 + 2x^4 + x^5.
c) The real roots of the polynomial is calculated by the roots function in MATLAB.
MATLAB output:
roots(coeff)
ans =
-1.8165 + 0.0000i
0.0000 + 0.5000i
0.0000 - 0.5000i

10MA209 HOMEWORK 2
0.2500 + 0.0000i
-0.1835 + 0.0000i
Hence, the real roots are x1 = -1.8165, x2 = 0.25, x3 = -0.1835.