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

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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