THE UNIVERSITY OF AUCKLAND
ENGSCI 211
SEMESTER ONE 2021 Campus: City
MATHEMATICAL MODELLING 2
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Page 1 of 7
Question 1 (9 marks)
Consider the ordinary differential equation (ODE):
y′′ +4y′ +5y = 2te−t
subject to the initial conditions:
y(0)=0, y′(0)=0
(a) Find the complementary function, yc, for this ODE. Write in a trigonometric form if
appropriate.
(b) Find the particular integral, yp.
ENGSCI 211
Trial solution, yC = Aeλt, substituted into homogeneous form of the ODE: λ2 + 4λ + 5 = 0 √
λ=−4± 16−20 22
λ = −2 ± i
Which gives a complementary function in trigonometric form:
yc =e−2t(Acos(t)+Bsin(t))
Trial solution (no resonance) will be of a similar form (i.e. product of exponential and polynomial):
PI001 Substituting back into the ODE:
yP = b0te−t + b1e−t
y′ =b e−t −b te−t −b e−t
y′′ = −2b e−t +b te−t −b e−t
te−t(b0 −4b0 +5b0)+e−t(−2b0 +b1 +4b0 −4b1 +5b1)=2te−t Comparing similar terms:
te−t : 2b0=2
e−t : 2b0+2b1=0
which gives b0 = 1 and b1 = −1. The particular integral is therefore: yp = te−t − e−t
(c) Find the total solution to the ODE.
Page 2 of 7
ENGSCI 211
Total solution, y = yc + yp, and its derivative: y=e−2t(Acos(t)+Bsin(t))+te−t −e−t
y′ =e−2t((−2A+B)cos(t)+(−A−2B)sin(t))+2e−t −te−t Applying initial conditions:
0 = −2A + B + 2
Solving gives A = 1 and B = 0. Total solution is given by: y=e−2tcos(t)+te−t −e−t
Page 3 of 7
Question 2 (9 marks)
(a) Consider the ordinary differential equation (ODE):
y′′ + 3y′ + 2y = 2t
subject to the initial conditions:
y(0)=1, y′(0)=0
Write the ODE and its initial conditions as a system of first-order ODEs.
(b) Perform two iterations of the Euler method, for a time step of ∆t = 1, for the above system of first-order ODEs. Use this to write the numerical estimate of y(2). (6 marks)
ENGSCI 211
Variable substitution:
System of ODEs:
yB′ =2t−2yA −3yB
yA(0)=1 yB (0)=0
yA = y yB =y′
1 01 0 −2 −2
1 −2 −1 −2 6 4
11 Therefore, Euler method is y(2) ≈ −1.
Page 4 of 7
Question 3 (3 marks)
The Laplace transform of an ODE has given the following expression for Y (s): e−s s−1 4
Y =s2+1+(s−1)2+1+s3
Find the ODE solution by taking the inverse Laplace transform of Y (s).
ENGSCI 211
Through application of linearity theorem:
−1 e−s −1 s − 1 −1 4
y=L s2 +1 +L (s−1)2 +1 +L s3
Now we can find each of these inverse terms, which requires use of different theorems. The first
term can be found using shift-in-t theorem:
f(t)=L−1 1
L s2 +1 =u(t−1)f(t−1) −1e−s
L s2 +1 =u(t−1)sin(t−1) The second term can be found using shift-in-s theorem:
−1 s−1 t −1 s L (s − 1)2 + 1 = e L s2 + 1
−1 s−1 t
L (s − 1)2 + 1 = e cos (t)
f (t) = sin (t) −1e−s
The third term does not require any special theorem:
−1 4 −1 2
The ODE solution is therefore given by:
y = u (t − 1) sin (t − 1) + et cos (t) + 2t2
L s3 =2L s3 −1 4 2
Page 5 of 7
Question 4 (7 marks)
The temperature of a metal sheet is given by:
T(x,y)=x3 −12x+y3 +3y2 −9y+20
Find and classify the stationary points of the temperature.
ENGSCI 211
First-order derivatives:
∂T = 3×2 − 12 ∂x
∂T = 3y2 + 6y − 9 ∂y
The first partial derivative is equal to zero for x = ±2 and any y value. The second partial derivative is equal to zero for y = −1 ± 2 and any x value.
The stationary points are therefore at: (−2, −3), (−2, 1), (2, −3) and (2, 1).
The Hessian matrix is given by:
6x 0 H= 0 6y+6
Classifying each stationary point using the Hessian matrix:
Point H det (H) ∂2T Classification ∂x2
(−2, −3) (−2, 1) (2, −3) (2, 1)
−12 0 0 −12
−12 0 0 12
12 0 0 −12
12 0 0 12
>0 <0 <0 >0
local maximum saddle point saddle point local minimum
Page 6 of 7
Question 5 (7 marks)
Consider the double integral
√R whereRistheregionboundedbyy= x,y=2,andx=0.
(a) Sketch the region of integration.
siny3 dA
(b) Write the double integral as an integral of the form f(x,y) dx dy (i.e. with x as the variable of the inner integral), and clearly identify the bounds of integration. (3 marks)
ENGSCI 211
−1 12345 −1
siny3 dxdy
(c) Hence evaluate the integral.
siny3 dxdy
= xsiny0dy
= −13 (cos(8) − 1)
y2siny3 dy 1 32
Page 7 of 7
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