SME
Differential Equations · Problem type
Airy’s equation by a power series
Use this for “Find the first terms of the series solution of y″ = xy with y(0) = 1, y′(0) = 0, and estimate y(1).”
Chart
Starting from y(0) = 1, y′(0) = 0; solid: worked out numerically
Dashed: the degree-6 Taylor polynomial about x = 0
At x = 1: P = 1.1722, y = 1.1723, error |y − P| = 7.775 × 10⁻⁵
This is an example. Type your own values to start a new problem.
y″
=
xy
y(0)
=
a₀
,
y′(0)
=
a₁
S
(
x
)
=
S
y(x)
Numerical solution at x
worked out (not typed)
E
Error |y(x) − S|
worked out (not typed)
Formulas
S
=
the
series
of
y″
=
xy
from
y(0)
=
a₀
,
y′(0)
=
a₁
,
to
degree
N
,
at
x
1.1722 = the series of y″ = xy from y(0) = 1, y′(0) = 0, to degree 6, at 1
y(
x
)
=
the
solution
of
y″
=
xy
from
y(0)
=
a₀
,
y′(0)
=
a₁
,
at
x
1.1723 = the solution of y″ = xy from y(0) = 1, y′(0) = 0, at 1
E
=
|y(
x
)
−
S
|
7.7748 ×
10
-5
=
|1.1723
−
1.1722|
Assumptions
•
Put y = Σ aₙxⁿ into y″ = xy. The x on the right raises each power by one, so 2a₂ = 0 and aₙ₊₂ = aₙ₋₁ ÷ ((n + 2)(n + 1)).
•
Each chain steps by three powers: a₀ = 1 gives 1 + x³/6 + x⁶/180 + …, and a₁ = 1 gives x + x⁴/12 + ….
•
No formula in elementary functions solves it, so y(x) here comes from a careful numerical solution.
Step-by-step
WE KNOW
a₀ = 1, a₁ = 0, N = 6, x = 1
FIND
partial sum up to xᴺ (S), numerical solution at x (y(x)), error |y(x) − S| (E)
Step 1 · Find partial sum up to xᴺ (S)
Use S = the series of y″ = xy from y(0) = a₀, y′(0) = a₁, to degree N, at x
a₂
=
0
and
aₙ₊₂
=
aₙ₋₁
÷
((n
+
2)(n
+
1))
from
a₀
and
a₁
:
a₂
=
0,
a₃
=
0.1667,
a₄
=
0,
a₅
=
0,
and
so
on
to
a₆.
Each
coefficient
times
its
power
of
x
,
added
up.
S
=
1
+
0.1667
×
x
3
+
0.005556
×
x
6
S
=
1
+
0.1667
×
1
3
+
0.005556
×
1
6
S
=
1
+
0.1666667
+
0.005555556
=
1.1722
S
=
1.1722
Step 2 · Find numerical solution at x (y(x))
Use y(x) = the solution of y″ = xy from y(0) = a₀, y′(0) = a₁, at x
Worked
out
numerically
in
steps
of
0.005
from
x
=
0,
to
compare
the
partial
sum
with.
y(
x
)
=
1.1723
Step 3 · Find error |y(x) − S| (E)
Use E = |y(x) − S|
How far the partial sum is from the solution: the terms after xᴺ, left out.
E
=
|1.17229997
−
1.17222222|
E
=
7.7748 ×
10
-5
Check
Put the numbers back in. Both sides should match.
1.1722
=
1
+
0.1667
×
1
3
+
0.005556
×
1
6
✓
1.1723 = the solution of y″ = xy from y(0) = 1, y′(0) = 0, at 1 ✓
7.7748 ×
10
-5
=
|1.1723
−
1.17222|
✓
Topic check
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