Linear Ordinary Differential Equation, Prüfungen von Mathematik

A linear differential equation or a system of linear equations such that the associated homogeneous equations have constant coefficients may be solved by quadrature, which means that the solutions may be expressed in terms of integrals. This is also true for a linear equation of order one, with non-constant coefficients. An equation of order two or higher with non-constant coefficients cannot, in general, be solved by quadrature. For order two, Kovacic's algorithm allows deciding whether there are solutions in terms of integrals, and computing them if any.

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UNIVERSITY OF ZIMBABWE
HASTS114
HACS113
HFM113
HDSCS116
BSc Honours in Mathematics Level 2
LINEAR MATHEMATICS AND ORDINARY DIFFERENTIAL EQUATIONS
July 2022
Time : 2 hours
Candidates may attempt a total of FIVE questions, at MOST TWO questions from each
section.
SECTION A
A1. Determine whether the vector y=
5
3
0
is a linear combination of the vectors
y1=
1
2
1
,y2=
3
1
2
,y3=
1
5
4
,y4=
6
5
1
. [10]
A2.
Use determinants/row reduction to solve the following system of linear equations
2x1+x2+ 7x32x4= 4
3x12x2+ 11x4= 13
x1+x2+ 5x33x4= 1.
[10]
A3. (a) Find all solutions to the following system of linear equations
2x13x2+x3+ 7x4= 14
2x1+ 8x24x3+ 5x4=1
x1+ 3x23x3= 4
5x1+ 2x2+ 3x3+ 4x4=19.
[10]
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UNIVERSITY OF ZIMBABWE

HASTS

HACS

HFM

HDSCS

BSc Honours in Mathematics Level 2 LINEAR MATHEMATICS AND ORDINARY DIFFERENTIAL EQUATIONS

July 2022 Time : 2 hours

Candidates may attempt a total of FIVE questions, at MOST TWO questions from each section.

SECTION A

A1. Determine whether the vector y =

 (^) is a linear combination of the vectors

y 1 =

 (^) , y 2 =

 (^) , y 3 =

 (^) , y 4 =

. [10]

A2.

Use determinants/row reduction to solve the following system of linear equations 2 x 1 + x 2 + 7x 3 − 2 x 4 = 4 3 x 1 − 2 x 2 + 11x 4 = 13 x 1 + x 2 + 5x 3 − 3 x 4 = 1. [10]

A3. (a) Find all solutions to the following system of linear equations

2 x 1 − 3 x 2 + x 3 + 7x 4 = 14 2 x 1 + 8x 2 − 4 x 3 + 5x 4 = − 1 x 1 + 3x 2 − 3 x 3 = 4 − 5 x 1 + 2x 2 + 3x 3 + 4x 4 = − 19. [10] page 1 of 3

HASTS114/HACS113/HFM113/HDSCS

(b) If V is any vector space, (i) Define W, a subspace of V. (ii) Define S, a basis for V. [2,3] (c) Find bases for the eigenspaces of  

[10]

END OF QUESTION PAPER

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