You are required to find the numerical solution to the following ordinary differential equation d²y dy + dx² dx in the domain 0≤x≤ 1. You are required to solve this equation using numerical methods with the boundary conditions y(0) = -1 and y(1) = 2. Note that the input for the sin term should be in radians (not degrees). As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes dy d²y dx² dx d²y dx² -6y = 15 sin(12x) (xi)~ (Yi+1 − Yi-1)/(2A) dy dx dy -(N-1)~ (YN-1 - YN-2)/A. dx The double derivative can be approximated as d²y dx² (xo)~ (y₁ - yo)/A (xi) ~ (Yi-1 - 2yi + Yi+1)/A² (xo) ≈ (yo - 2y1 + y2)/A² (N-1)~ (YN-1-2YN-2+ YN-3)/A². Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as

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You are required to find the numerical solution to the following ordinary differential equation
d²y dy
+
dx² dx
in the domain 0 ≤ x ≤ 1. You are required to solve this equation using numerical methods with
the boundary conditions y(0) = −1 and y(1) = 2. Note that the input for the sin term should be
in radians (not degrees).
As shown in lectures, the derivative of a function y(x) can be calculated using the second order
central, and first order forward and backward difference schemes
d²y
dx²
dy
- (xi) ≈ (Yi+1 − Yi-1)/(2A)
dx
d²y
6y = 15 sin(12x)
dy
-(XN−1) ≈ (YN-1 — YN-2)/▲.
dx
The double derivative can be approximated as
dx²
dy
d²y
· (xo) ≈ (y₁ − Yo)/A
dx
; (x;) ≈ (Yi−1 − 2Yi + Yi+1)/A²
dx²
; (xo) ~ (yo − 2y₁ +Y2)/▲²
(N-1) ~ (YN-1 — 2YN−2+ YN-3)/A².
Show that if you discretise the equation shown in Q1 using the differentiation schemes above and
implementing the boundary conditions, you will get a set of equations that can be expressed as
Transcribed Image Text:You are required to find the numerical solution to the following ordinary differential equation d²y dy + dx² dx in the domain 0 ≤ x ≤ 1. You are required to solve this equation using numerical methods with the boundary conditions y(0) = −1 and y(1) = 2. Note that the input for the sin term should be in radians (not degrees). As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes d²y dx² dy - (xi) ≈ (Yi+1 − Yi-1)/(2A) dx d²y 6y = 15 sin(12x) dy -(XN−1) ≈ (YN-1 — YN-2)/▲. dx The double derivative can be approximated as dx² dy d²y · (xo) ≈ (y₁ − Yo)/A dx ; (x;) ≈ (Yi−1 − 2Yi + Yi+1)/A² dx² ; (xo) ~ (yo − 2y₁ +Y2)/▲² (N-1) ~ (YN-1 — 2YN−2+ YN-3)/A². Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as
where
1
α1
0
0
0
0
0
000
B₁
a2 B₂
0
of ♂ …..ooo
::0
0
Y1
0
0 a3 B3
೪೦೦
0
0
0
72
0
0
0
Y3
0000
aN-3 BN-3 YN-3
0
=
:
0
aN-2 BN-2 YN-2
0
0
1
di
Bi
Vi
Qi
=
=
A²
2
4²
0
0
0
2A
-6
1
2A
Δ2
15 sin(12x₁)
a = -1, b = 2
S5
Yo
Y1
Y2
Y3
YN-3
YN-2
YN-1
||
a
Q₁
Q2
Q3
:
QN-3
QN-2
b
and Xi+1
= xį + A where A is the spacing between your equally spaced grid points. There are a
total of N grid points. Note that co
0 and N-1
=
1 and the notation yi is the approximate
value of y(x₂).
Transcribed Image Text:where 1 α1 0 0 0 0 0 000 B₁ a2 B₂ 0 of ♂ …..ooo ::0 0 Y1 0 0 a3 B3 ೪೦೦ 0 0 0 72 0 0 0 Y3 0000 aN-3 BN-3 YN-3 0 = : 0 aN-2 BN-2 YN-2 0 0 1 di Bi Vi Qi = = A² 2 4² 0 0 0 2A -6 1 2A Δ2 15 sin(12x₁) a = -1, b = 2 S5 Yo Y1 Y2 Y3 YN-3 YN-2 YN-1 || a Q₁ Q2 Q3 : QN-3 QN-2 b and Xi+1 = xį + A where A is the spacing between your equally spaced grid points. There are a total of N grid points. Note that co 0 and N-1 = 1 and the notation yi is the approximate value of y(x₂).
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