Second Order ODE/y'' - x f(x) y' + f(x) y = 0

Theorem

The second order ODE:

$(1): \quad y' ' - x \, \map f x y' + \map f x y = 0$

has the general solution:

$\ds y = C_1 x + C_2 x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x$

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

Substituting into $(1)$:

\(\ds y' ' - x \map f x y' + \map f x y\)

\(=\)

\(\ds 0 - x \map f x 1 + \map f x x\)

\(\ds \)

\(=\)

\(\ds 0\)

and so it has been demonstrated that:

$y_1 = x$

is a particular solution of $(1)$.

$(1)$ is in the form:

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

where:

$\map P x = -x \map f x$

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:

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

\(=\)

\(\ds \int \paren {-x \map f x} \rd x\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds e^{-\int \paren {-x \map f x} \rd x}\)

\(\ds \)

\(=\)

\(\ds e^{\int x \map f x \rd x}\)

Hence:

\(\ds v\)

\(=\)

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

Definition of $v$

\(\ds \)

\(=\)

\(\ds \int \dfrac 1 {x^2} e^{\int x \map f x \rd x} \rd x\)

as $y_1 = x$

and so:

\(\ds y_2\)

\(=\)

\(\ds v y_1\)

Definition of $y_2$

\(\ds \)

\(=\)

\(\ds x \int x^{-2} e^{\int x \map f x \rd x} \rd x\)

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

$\ds y = C_1 x + C_2 x \int x^{-2} e^{\int x \map f x \rd x} \rd x$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $8$