Fall 2019-20 MAT2002: Differential Equations Assignment Solutions

This document presents solutions to a homework assignment on the applications of differential and difference equations (MAT2002). The assignment covers two main experiments: solving homogeneous systems of first and second-order differential equations using matrix methods, and finding series solutions of ordinary differential equations. The solutions include detailed MATLAB code, input/output examples, and explanations for each step. The first part focuses on solving systems of differential equations with given initial conditions, while the second part focuses on finding the first five terms of power series solutions for differential equations, also with given initial conditions. The document is designed to help students understand and solve complex differential equation problems using computational tools.

Experiment:4A
Solution of homogeneous system of first order
and second order differential equations by matrix
method
QUESTION 1:
Solve the system of differential equations y0 1 = 4y1 +
y2 , y0 2 = 3y1 + 2y2, with the initial conditions y1(0)
= 2, y2(0) = 0.
AIM:
finding solution of given differential equation with the
help of given intial conditions given
MATLAB CODE:
clc
clear all
close all
syms t c1 c2
A=input('Enter a square matrix :');
[v,d]=eig(A)
y1=c1*exp(d(1)*t)
y2=c2*exp(d(4)*t)
X=v*[y1;y2]
IC=input('Enter ICs in the
form[t0,x1(t0),x2(t0)] :');
eq1=subs(X(1),IC(1))-IC(2);
eq2=subs(X(2),IC(1))-IC(3);
[c1,c2]=solve(eq1,eq2);
X=subs(X)
Applications of Differential and Difference Equations(MAT2002)
Instructor: Dr. Aruna. KFall Semester2019-20
Department of Mathematics, School of advanced sciences

OUTPUT:
X =
(3*exp(5*t))/2 + exp(t)/2
(3*exp(5*t))/2 - (3*exp(t))/2
QUESTION 2 :
Solve the system of differential equation y’’1 = 2y1+y2, y’’2 =
y1+2y2, with the initial conditions y1(0) = 0,y’1(0) = 1, y2(0) =
1,y’2(0) = 0.

AIM:
To find the functions of y1(x) and y2(x)
MATLAB CODE:
clc
clear all
close all
syms t x1(t) x2(t) Dx1(t) Dx2(t)
A=input('Enter a square matrix :');
[p,d]=eig(A);
eq1=dsolve(['D2x1=',num2str(d(1)),'*x1']);
eq2=dsolve(['D2x2=',num2str(d(4)),'*x2']);
X=[eq1;eq2]
X=p*X
OUTPUT:
X =
(2^(1/2)*(C1*exp(3^(1/2)*t) + C2*exp(-
3^(1/2)*t)))/2 - (2^(1/2)*(C2*exp(t) +
C1*exp(-t)))/2
(2^(1/2)*(C2*exp(t) + C1*exp(-t)))/2 +
(2^(1/2)*(C1*exp(3^(1/2)*t) + C2*exp(-
3^(1/2)*t)))/2
Experiment: 4B
Series solutions of ordinary differential equations

QUESTION 1 :
Find the first five terms in the power series solution of
the differential equation y’’+ x2 * y = 0 with the initial
conditions y(0) = 1, y’(0) = 2
AIM:
To find the power series of the given differential
equations
MATLAB CODE:
clc
clear all
close all
syms x a0 a1 a2 a3 a4 a5 A B t
p=input('enter p:');
q=input('enter q:');
r=input('enter r:');
z=input('enter point:')
a = [a0 a1 a2 a3 a4 a5]
y = sum(a.*(x-z).^[0:5])
dy = diff(y)
d2y = diff(dy)
de = collect(p*d2y+q*dy+r*y,(x-z))
de=subs(de,x-z,t)
coef=coeffs(de,t);
A2=solve(coef(1),a2)
A3=solve(coef(2),a3)
A4=subs(solve(coef(3),a4),a2,A2)
A5=subs(solve(coef(4),a5),{a2,a3},{A2,A3})
y=subs(y,{a2,a3,a4,a5},{A2,A3,A4,A5})
soln=coeffs(y,[a1 a0])
gs=A*soln(1)+B*soln(2)
IC=input('Enter ICs in vector form[x y(x)
dy(x)]: ');
eq1=subs(gs,x,IC(1))-IC(2);
eq2=subs(diff(gs),x,IC(1))-IC(3);
[A B]=solve(eq1,eq2)

G_S=subs(gs)
OUTPUT:
G_S =
- x^5/10 - x^4/12 + 2*x + 1