Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0/Proof 1

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Theorem

The second order ODE:

$(1): \quad x^2 y'' + x y' - 4 y = 0$

has the general solution:

$y = C_1 x^2 + \dfrac {C_2} {x^2}$


Proof

Note that:

\(\ds y_1\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds 2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds y''\) \(=\) \(\ds 2\) Power Rule for Derivatives

and so by inspection:

$y_1 = x^2$

is a particular solution of $(1)$.


$(1)$ can be expressed as:

$(2): \quad y'' + \dfrac 1 x y' - \dfrac 4 {x^2} y = 0$

which is in the form:

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

where:

$\map P x = \dfrac 1 x$
$\map Q x = -\dfrac 4 {x^2}$


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:

\(\ds \int P \rd x\) \(=\) \(\ds \int \dfrac 1 x \rd x\)
\(\ds \) \(=\) \(\ds \ln x\) Primitive of Reciprocal
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{-\ln x}\)
\(\ds \) \(=\) \(\ds \frac 1 x\)


Hence:

\(\ds v\) \(=\) \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) Definition of $v$
\(\ds \) \(=\) \(\ds \int \dfrac 1 {x^4} \frac 1 x \rd x\)
\(\ds \) \(=\) \(\ds \int \dfrac 1 {x^5} \rd x\)
\(\ds \) \(=\) \(\ds - \frac 1 {4 x^4}\)


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds -\frac 1 {4 x^4} x^2\)


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

$y = C_1 x^2 + k \paren {-\dfrac 1 {4 x^2} }$

where $k$ is arbitrary.


Setting $C_2 = - \dfrac k 4$ yields the result:

$y = C_1 x^2 + \dfrac {C_2} {x^2}$

$\blacksquare$


Sources