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Euler Differential Equation

The general nonhomogeneous equation is

x^2 {d^2y\over dx^2} + \alpha x {dy\over dx} + \beta y = S(x).
\end{displaymath} (1)

The homogeneous equation is
x^2y''+\alpha xy'+\beta y = 0
\end{displaymath} (2)

y'' + {\alpha\over x} y' + {\beta\over x^2} y = 0.
\end{displaymath} (3)

Now attempt to convert the equation from
\end{displaymath} (4)

to one with constant Coefficients
{d^2y\over dz^2} + A {dy\over dz} + By = 0
\end{displaymath} (5)

by using the standard transformation for linear Second-Order Ordinary Differential Equations. Comparing (3) and (5), the functions $p(x)$ and $q(x)$ are
p(x) \equiv {\alpha\over x} = \alpha x^{-1}
\end{displaymath} (6)

q(x) \equiv {\beta\over x^2} = \beta x^{-2}.
\end{displaymath} (7)

Let $B\equiv\beta$ and define
$\displaystyle z$ $\textstyle \equiv$ $\displaystyle B^{-1/2}\int \sqrt{q(x)}\,dx = \beta^{-1/2} \int\sqrt{\beta x^{-2}}\,dx$  
  $\textstyle =$ $\displaystyle \int x^{-1}\,dx = \ln x.$ (8)

Then $A$ is given by
$\displaystyle A$ $\textstyle \equiv$ $\displaystyle {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2}$  
  $\textstyle =$ $\displaystyle {-2\beta x^{-3}+2(\alpha x^{-1})(\beta x^{-2})\over 2(\beta x^{-2})^{3/2}} \beta^{1/2}$  
  $\textstyle =$ $\displaystyle \alpha-1,$ (9)

which is a constant. Therefore, the equation becomes a second-order ODE with constant Coefficients
{d^2y\over dz^2}+(\alpha-1){dy\over dz}+\beta y = 0.
\end{displaymath} (10)

$\displaystyle r_1$ $\textstyle \equiv$ $\displaystyle {\textstyle{1\over 2}}\left({-A+\sqrt{A^2-4B}\,}\right)$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\left[{1-\alpha+\sqrt{(\alpha-1)^2-4\beta}\,}\right]$ (11)
$\displaystyle r_2$ $\textstyle \equiv$ $\displaystyle {\textstyle{1\over 2}}\left({-A-\sqrt{A^2-4B}\,}\right)$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\left[{1-\alpha-\sqrt{(\alpha-1)^2-4\beta}\,}\right]$ (12)

$\displaystyle a$ $\textstyle \equiv$ $\displaystyle {\textstyle{1\over 2}}(1-\alpha )$ (13)
$\displaystyle b$ $\textstyle \equiv$ $\displaystyle {\textstyle{1\over 2}}\sqrt{4\beta -(\alpha -1)^2}.$ (14)

The solutions are
$\displaystyle y$ $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} c_1e^{r_1z}+c_2e^{r_2z} & \mbox{$(\alpha...
...{az}[c_1\cos(bz)+c_2\sin(bz)] & \mbox{$(\alpha-1)^2<4\beta$.}\end{array}\right.$  

In terms of the original variable $x$,

$\displaystyle y$ $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} c_1\vert x\vert^{r_1}+c_2\vert x\vert^{r...
...t)+c_2\sin(b\ln\vert x\vert)] & \mbox{$(\alpha-1)^2<4\beta$.}\end{array}\right.$ (16)

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© 1996-9 Eric W. Weisstein