Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0

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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