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

Theorem

The second order ODE:

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

has the general solution:

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

Proof

The particular solution:

$y_1 = x$

can be found by inspection.

Let $(1)$ be written as:

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

which is in the form:

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

where:

$\map P x = \dfrac 1 x$

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

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:

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

\(=\)

\(\ds \int \frac 1 x \rd x\)

\(\ds \)

\(=\)

\(\ds \ln x\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds e^{-\ln x}\)

\(\ds \)

\(=\)

\(\ds \frac 1 x\)

Hence:

\(\ds v\)

\(=\)

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

Definition of $v$

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \int \dfrac {\d x} {x^3}\)

\(\ds \)

\(=\)

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

and so:

\(\ds y_2\)

\(=\)

\(\ds v y_1\)

Definition of $y_2$

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds -\dfrac 1 {2 x}\)

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

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

where $k$ is arbitrary.

Setting $C_2 = -\dfrac k 2$ yields the result:

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

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: The Use of a Known Solution to find Another: Example $1$