Legendre's Differential Equation/(1 - x^2) y'' - 2 x y' + 2 y = 0

Theorem

The special case of Legendre's differential equation:

$(1): \quad \paren {1 - x^2} y - 2 x y' + 2 y = 0$

has the general solution:

$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$

Proof

Note that:

\(\ds y_1\)

\(=\)

\(\ds x\)

\(\ds \leadsto \ \ \)

\(\ds {y_1}'\)

\(=\)

\(\ds 1\)

Power Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds {y_1}\)

\(=\)

\(\ds 0\)

Derivative of Constant

and so by inspection:

$y_1 = x$

is a particular solution of $(1)$.

$(1)$ can be expressed as:

$(2): \quad y - \dfrac {2 x} {1 - x^2} y' + \dfrac 2 {1 - x^2} y = 0$

which is in the form:

$y + \map P x y' + \map Q x y = 0$

where:

$\map P x = - \dfrac {2 x} {1 - x^2}$

$\map Q x = \dfrac 2 {1 - x^2}$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.

We have that:

\(\ds \int P \rd x\)

\(=\)

\(\ds \int \paren {-\dfrac {2 x} {1 - x^2} } \rd x\)

\(\ds \)

\(=\)

\(\ds \map \ln {1 - x^2}\)

Primitive of Function under its Derivative

\(\ds \leadsto \ \ \)

\(\ds e^{-\int P \rd x}\)

\(=\)

\(\ds e^{-\map \ln {1 - x^2} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 {1 - x^2}\)

Hence:

\(\ds v\)

\(=\)

\(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\)

Definition of $v$

\(\ds \)

\(=\)

\(\ds \int \dfrac 1 {x^2} \frac 1 {1 - x^2} \rd x\)

\(\ds \)

\(=\)

\(\ds \int \dfrac 1 {x^2 \paren {1 - x^2} } \rd x\)

\(\ds \)

\(=\)

\(\ds -\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\)

Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2} }$

and so:

\(\ds y_2\)

\(=\)

\(\ds v y_1\)

Definition of $y_2$

\(\ds \)

\(=\)

\(\ds \paren {-\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } } x\)

\(\ds \)

\(=\)

\(\ds -1 + \frac x 2 \map \ln {\frac {1 + x} {1 - x} }\)

From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $5$