Euler Differential Equation

The general nonhomogeneous equation is

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

The homogeneous equation is

$$x^2y''+\alpha xy'+\beta y = 0$$ (2)

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

Now attempt to convert the equation from

$$y''+p(x)y'+q(x)y=0$$ (4)

to one with constant Coefficients

$${d^2y\over dz^2} + A {dy\over dz} + By = 0$$ (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

$$p(x) \equiv {\alpha\over x} = \alpha x^{-1}$$ (6)

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

Let $B\equiv\beta$ and define

$$z \equiv B^{-1/2}\int \sqrt{q(x)}\,dx = \beta^{-1/2} \int\sqrt{\beta x^{-2}}\,dx$$

$$= \int x^{-1}\,dx = \ln x.$$ (8)

Then $A$ is given by

$$A \equiv {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2}$$

$$= {-2\beta x^{-3}+2(\alpha x^{-1})(\beta x^{-2})\over 2(\beta x^{-2})^{3/2}} \beta^{1/2}$$

$$= \alpha-1,$$ (9)

which is a constant. Therefore, the equation becomes a second-order ODE with constant Coefficients

$${d^2y\over dz^2}+(\alpha-1){dy\over dz}+\beta y = 0.$$ (10)

Define

$$r_1 \equiv {\textstyle{1\over 2}}\left({-A+\sqrt{A^2-4B}\,}\right)$$

$$= {\textstyle{1\over 2}}\left[{1-\alpha+\sqrt{(\alpha-1)^2-4\beta}\,}\right]$$ (11)

$$r_2 \equiv {\textstyle{1\over 2}}\left({-A-\sqrt{A^2-4B}\,}\right)$$

$$= {\textstyle{1\over 2}}\left[{1-\alpha-\sqrt{(\alpha-1)^2-4\beta}\,}\right]$$ (12)

and

$$a \equiv {\textstyle{1\over 2}}(1-\alpha )$$ (13)

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

The solutions are

$$y = \left\{\begin{array}{ll} c_1e^{r_1z}+c_2e^{r_2z} & \mbox{$(\alpha... ...{az}[c_1\cos(bz)+c_2\sin(bz)] & \mbox{$(\alpha-1)^2<4\beta$.}\end{array}\right.$$ (15)

In terms of the original variable $x$,

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