Second Order ODE/x y'' + 3 y' = 0
Jump to navigation
Jump to search
Theorem
The second order ODE:
- $(1): \quad x y'' + 3 y' = 0$
has the general solution:
- $y = C_1 + \dfrac {C_2} {x^2}$
Proof
Note that:
\(\ds y_1\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y'\) | \(=\) | \(\ds 0\) | Derivative of Constant | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y''\) | \(=\) | \(\ds 0\) | Derivative of Constant |
and so by inspection:
- $y_1 = 1$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
- $(2): \quad y'' + \dfrac 3 x y' = 0$
which is in the form:
- $y'' + \map P x y' + \map Q x y = 0$
where:
- $\map P x = \dfrac 3 x$
- $\map Q x = 0$
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:
\(\ds \int P \rd x\) | \(=\) | \(\ds \int \dfrac 3 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \ln x\) | Primitive of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \ln x^3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{-\int P \rd x}\) | \(=\) | \(\ds e^{-\ln x^3}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {x^3}\) |
Hence:
\(\ds v\) | \(=\) | \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) | Definition of $v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac 1 {x^3} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 x^2}\) |
and so:
\(\ds y_2\) | \(=\) | \(\ds v y_1\) | Definition of $y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 x^2}\) |
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $y = C_1 + k \paren {-\frac 1 {2 x^2} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 2$ yields the result:
- $y = C_1 + \dfrac {C_2} {x^2}$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $3$