Approximation Methods and Solutions of Differential Equations

This document presents a comprehensive solution to a differential equations assignment, addressing various problem-solving techniques. The assignment covers sketching graphs, estimating solutions using graphical methods, and analyzing data related to tidal heights and viscosity. It delves into numerical integration, calculating roots through iterative techniques such as Newton's and Regula Falsi methods, and solving differential equations related to damped mechanical systems and electrical circuits. The solutions involve applying concepts of integration, differentiation, and algebraic manipulation to arrive at the solutions. This assignment serves as a valuable resource for students studying calculus and differential equations, offering detailed explanations and step-by-step solutions to enhance understanding and problem-solving skills. The content is designed to aid students in grasping the core principles of differential equations and their practical applications.

Differential equations
approximation methods

Table of Contents
TASK 2............................................................................................................................................1
2.1 & 2.2......................................................................................................................................1
2.3 Numerical integration............................................................................................................8
4.1 (part), 4.2 (part) & 4.5 (part)....................................................................................................12
4.1 (part), 4.2 (part) & 4.5 (part)....................................................................................................13
(b)...............................................................................................................................................13
c)................................................................................................................................................14
4.3 (part), 4.4 & 4.5 (part)..............................................................................................................14
(a)...............................................................................................................................................14
b)................................................................................................................................................16
(c)...............................................................................................................................................16

TASK 2
2.1 & 2.2
(a) Sketch of following graphs -
i) y = 2 – 3 cos (0.5x – π/5)
at x = 100,
y = 2 – 3 cos (50 - π/5)
= 2 – 3 (0.9)
= 2 – 2.7 = -0.7
at x = 200
y = 2 – 3 cos (100 – π/5)
= 2 – 3 (0.4)
= 2 – 1.2 = 0.8
at x = 300
y = 2 – 3 cos (150 – π/5)
= 2 – 3 (-0.4)
= 2 + 1.2 = 3.2
x y
100 -0.7
200 0.8
300 3.2
400 3.8
1

ii) y = 0.5x3 – 1.9x2 – 12.12x + 33.12
at x = 0,
y = 0.5(0)3 – 1.9(0)2 – 12.12(0) + 33.12
= 33.12
at x = 1,
y = 0.5(1)3 – 1.9(1)2 – 12.12(1) + 33.12
= 0.5 – 1.9 – 12.12 + 33.12
= 19.6
at x = -1,
y = 0.5(-1)3 – 1.9(-1)2 – 12.12(-1) + 33.12
= -0.5 – 1.9 + 12.12 + 33.12
= 42.84
x y
0 33.12
1 19.6
-1 42.84
2

(b) Estimated solutions using graphical method –
2x – 5 = -x2 – 4x + 5
-x + 3
On simplifying,
2x + 5 = (-x + 3) (-x2 – 4x + 5)
→ 2x + 5 = x3 + x2 – 17x + 15
→ x3 + x2 – 19x + 10 = 0
Through graph, it has estimated value of x ≈ -5.1, 0.55
3
y = 0.5x3 – 1.9x2 – 12.12x + 33.12

c)
Day Tidal
Height
1 0.499507
2 0.842754
3 1.809209
4 3.206897
5 4.758573
4

6 6.156701
7 7.124383
8 7.470172
9 7.153591
10 6.182957
11 4.780279
12 3.223236
13 1.819976
14 0.848065
5

d)
(i) Estimation of Value of viscosity at 35 C temperature through graph⁰
Through this graph, it has been evaluated that when temperature is 35 C, then estimated value of⁰
viscosity will be 0.85 cP
(ii) Estimation of Value of viscosity at 35 C temperature through linear relationship⁰
From above graph, the linear relationship between viscosity and temperature is depicted as –
y = -0.0178x + 1.6308
where, y represents viscosity and x as temperature,
then
at 35 C, Viscosity will be –⁰
y = -0.0178(35) + 1.6308
= 1.7008 cP
6

(iii) Estimation of temperature when viscosity is 1.6cP
Since linear relationship from graph between viscosity and temperature is evaluated as –
y = -0.0178x + 1.6308
where, y represents viscosity and x as temperature,
then
at 1.6cP, Temperature will be –
1.6 = -0.0178x + 1.6308
0.0178x = 1.6308 – 1.6
x = 0.0308/0.0178
= 1.7 C⁰
7

2.3 Numerical integration
a) 1.3
∫0.2 exp (-x)2 .dx
1.3
Let I = ∫0.2 exp (-x)2 .dx
1.3 2
I2 = ∫0.2 exp (-x)2 .dx
1.3 1.3
= ∫0.2 exp (-x)2 .dx ∫0.2 exp (-x)2 .dx
1.3 1.3
= ∫0.2 exp (-x)2 .dx ∫0.2 exp (-y)2 .dy
1.3 1.3
= ∫0.2 ∫0.2 exp (-x)2 .dx exp (-y)2 .dy
1.3 1.3
= ∫0.2 ∫0.2 exp -(x2 + y2).dx dy
Now, change the coordinated to polar form, as –
x2 + y2 = r2
which represents a circle,
then
2π 1.3
= ∫0. ∫0.2 exp -(r2 ).r dr dϴ
2π 1.3
= ∫0. [- ½ exp(-r2)]0.2dϴ
2π
= ∫0. [- ½ exp(-1.3) + ½ exp (0.2)]dϴ
8

2π
= ∫0. [- ½ exp(-1.3) + ½ exp (0.2)]dϴ
= - ½ [ϴ]02π
I2 = π
I = √π
9

2.4 Calculate roots by using iterative techniques
a) 2x – 5 = -x2 – 4x + 5
-x + 3
On simplifying,
2x + 5 = (-x + 3) (-x2 – 4x + 5)
→ 2x + 5 = x3 + x2 – 17x + 15
→ x3 + x2 – 19x + 10 = 0
Newton’s iterative method –
let y(x) = x3 + x2 – 19x + 10
y(-1) = 29
y(1) = -7
so, root of this equation lies between -1 and 1
First approximation
x1 = (-1+1)/2
= 0
so, f(0) = (0)3 + (0)2 – 19 (0) + 10
= 10
Hence, root lies between 0 and 1
Second approximation
x2 = (0+1)/2
= 0.5
so, f(0.5) = (0.5)3 + (0.5)2 – 19 (0.5) + 10
= 1.375
Hence, root lies between 0.5 and 1
Third approximation
x3 = (0.5 + 1)/2
= 0.75
so, f(0.75) = (0.75)3 + (0.75)2 – 19 (0.75) + 10
= 3.26 which is positive
Hence, root of the equation is approx. 0.75
10

(b) (x – 2)3 + 4cos x/3 – e0.4x = 2
On simplifying,
(x – 2)3 + 4cos x/3 – e0.4x – 2 = 0
x3 – 6x2 + 12x – 10 + 4cos x/3 – e0.4x = 0
Regula Falsi iterative method
let y(x) = x3 – 6x2 + 12x – 10 + 4cos x/3 – e0.4x
y(0) = -7
y(2) = 1.3
so, root of this equation lies between 0 and 2
Then, first approx.
x1= a f(b) – b f(a) /f(b) – f(a)
= 0 x (1.3) – 2 x (-7)/ 1.3 – (-7)
≈1.6
f(1.6) = (1.6)3 – 6(1.6)2 + 12(1.6) – 10 + 4cos 1.6/3 – e0.4(1.6)
= 1.6
which is positive, so root of the given equation is 1.6
11

4.1 (part), 4.2 (part) & 4.5 (part)
a) Given
dT = μT
dϴ
dT = μ dϴ
T
Integrating both side –
log T = μϴ
or, T = eμϴ
(i)
at ϴ = 2 radians
T = e2μ tension
12

4.1 (part), 4.2 (part) & 4.5 (part)
(b)
Given equation –
CR dV + V = E
dt
i) Solve equation for V at t = 0 and V = 0
Solution –
CR dV + V = E
dt
dV = 1 . dt
(E – V) CR
ln(E – V) = -t/CR + k
E – V = ke(-t/CR)
V = E - ke(-t/CR)
at initial condition, at V = 0 and t = 0
k = E
therefore,
V = E (1 – e(-t/CR))
ii) Calculate V when E = 25V, C = 20 μF, R = 200k Ω and t = 3.0s
Given, C = 20μF or, 20 x 10-6 F
and,
R = 200kΩ or 200 x 103Ω
Using, above equation –
V = E (1 – e(-t/CR))
= 25 (1 – e^(-3/20 x 10-6 . 200 x 103))
≈ 13.192
13

c)
Given,
dy = 3 – y
dx x
Let y = vx
then
dy/dx = v + x. dv/dx
so,
v + x. dv/dx = 3 – v
x. dv/dx = 3 – 2v
or,
dx = dv
x (3 – 2v)
integrating both side
∫ dx = ∫ dv
x (3 – 2v)
log x = - ½ log (3 – 2v)
or, x-2 = (3 – 2v)
x-2 = (3 – 2y/x)
3x2 – 2xy = 0
4.3 (part), 4.4 & 4.5 (part)
(a)
Displacement s of a body in a damped mechanical system –
2 d2s + 14ds + 20s = 0
dt2 dt
at t = 0, s = 0 and ds/dt = 3
Solution –
2 d2s + 14ds + 20s = 0
dt2 dt
14

Divide the given equation by 2,
d2s + 7ds + 10s = 0
dt2 dt
now, take D = d/dt
D2 + 7D + 10 = 0
(D + 5) (D + 2) = 0
D = -5, -2
Now, at s = 0, t = 0 and ds/dt = 3
So, s = Ae-5t + Be-2t (i)
differentiate –
ds/dt = -5A e-5t – 2 B e-2t …ii)
taking the given values s = 0, t = 0 and ds/dt = 3, in eq i) and eq ii)–
A + B = 0
and,
-5A – 2 B = 3
solving these two equation, A = 1 and B = -1
so,
[ s = -e-5t + e-2t ]
15

b)
Given –
L d2q + R dq + 1 q = E
dt2 dt C
where, value of L = 4H, C = 100μF = 100 x 10-6F, R = 400 and E = 240V
then,
Auxiliary equation is
Lm2 + Rm + 1/C = 0
then,
m = -R ± √R2 – 4L/C
2L
or, m = -50
so, general equation will be
q = (At + B) e-50t
(c)
5d2y + 9 dy – 2y = 3et
dt2 dt
solve y = 0.25, dy/dt = 0 at t = 0
Solution –
Since the general solution of
(5D2 + 9D – 2) y = 3et
substitute m for D
16

5m2 + 9m – 2 = 0
(5m – 1) (m + 2) = 0
or, m = 1,
then general equation will be
y = C1 et + C2 e-2t
While, particular solution will be –
y = Aet
so,
then y’ = Aet
and, y” = Aet
so,
(5Aet + 9Aet – 2Aet) = 3et
12Aet = 3et
or, A = ¼
so, Particular solution is y = ¼ et
17