When N is a positive integer, the Legendre equation ( 1 − x 2 ) y ″ − 2 x y ′ + N ( N + 1 ) y = 0 , With − 1 < x < 1 , has a solution that is a polynomial of degree N . Show by substitution into the differential equation that in the case N = 3 such a solution is y ( x ) = 1 2 x ( 5 x 2 − 3 )
When N is a positive integer, the Legendre equation ( 1 − x 2 ) y ″ − 2 x y ′ + N ( N + 1 ) y = 0 , With − 1 < x < 1 , has a solution that is a polynomial of degree N . Show by substitution into the differential equation that in the case N = 3 such a solution is y ( x ) = 1 2 x ( 5 x 2 − 3 )
Solution Summary: The author explains that y(x)=12x
When
N
is a positive integer, the Legendre equation
(
1
−
x
2
)
y
″
−
2
x
y
′
+
N
(
N
+
1
)
y
=
0
,
With
−
1
<
x
<
1
, has a solution that is a polynomial of degree
N
. Show by substitution into the differential equation that in the case
N
=
3
such a solution is
y
(
x
)
=
1
2
x
(
5
x
2
−
3
)
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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