Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0
Jump to navigation
Jump to search
Theorem
The second order ODE:
- $(1): \quad x y'' - \paren {2 x + 1} y' + \paren {x + 1} y = 0$
has the general solution:
- $y = C_1 e^x + C_2 x^2 e^x$
Proof
Note that:
- $x - \paren {2 x + 1} + \paren {x + 1} = 0$
so if $y'' = y' = y$ we find that $(1)$ is satisfied.
So:
\(\ds y_1\) | \(=\) | \(\ds e^x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_1}'\) | \(=\) | \(\ds e^x\) | Derivative of Exponential Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_1}''\) | \(=\) | \(\ds e^x\) | Derivative of Exponential Function |
and so:
- $y_1 = e^x$
is a particular solution of $(1)$.
$(1)$ can be expressed as:
- $(2): \quad y'' - \dfrac {2 x + 1} x y' + \dfrac {x + 1} x y = 0$
which is in the form:
- $y'' + \map P x y' + \map Q x y = 0$
where:
- $\map P x = -\dfrac {2 x + 1} x$
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:
\(\ds \int P \rd x\) | \(=\) | \(\ds \int \paren {-\dfrac {2 x + 1} x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {-2 - \dfrac 1 x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 x - \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{-\int P \rd x}\) | \(=\) | \(\ds e^{-\paren {-2 x - \ln x} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{2 x + \ln x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{2 x} e^{\ln x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x e^{2 x}\) |
Hence:
\(\ds v\) | \(=\) | \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) | Definition of $v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {e^{2 x} } x e^{2 x} \rd x\) | as $y_1 = e^x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2\) |
and so:
\(\ds y_2\) | \(=\) | \(\ds v y_1\) | Definition of $y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 e^x\) |
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $y = C_1 e^x + k \dfrac {x^2} 2 e^x$
and so setting $C_2 = \dfrac k 2$:
- $y = C_1 e^x + C_2 x^2 e^x$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $9$