[math]D<0[/math]Euler's Differential Equation is a regular differential equation with a singular point at [math]x=0[/math]. The second order form of the equation is [math]\text{x^2\,y'' + b\,x\,y' + c\,y = 0}[/math]. The solutions are found by substituting [math]y=x^r[/math] into the differential equation with [math]y'=r\,x^{r-1}[/math] and [math]y''=r\,(r-1)\,x^{r-2}[/math].[br]After simplifying a characteristic equation of [math]r^2+(b-1)\,r+c=0[/math] results. With a solution of [math]r = \frac{1-b \pm \sqrt{(b-1)^2 - 4\,c}}{2}[/math]. Then depending on the value inside the radical ,[math] D=(b-1)^2 - 4\,c[/math] the two solutions are:[br][math] D \gt 0 \::r_1 = \frac{1-b+\sqrt{D}}{2}\::\:y_1 = x^{r_1} [/math][br] [math]r_2 = \frac{1-b-\sqrt{D}}{2}\::\:y_2 = x^{r_2} [/math][br][math] D \lt 0 \::r = \frac{1-b}{2} \pm i \frac{\sqrt{D}}{2} = \lambda\pm i \mu\lambda \to y= x^\lambda(\alpha \cos( \mu\,\ln(x))+\beta \sin(\mu\ln(x)) [/math][br] [math]y_1 = x^\lambda(\alpha \cos( \mu\,\ln(x))[/math][br] [math]y_2 = x^\lambda(\alpha \sin( \mu\,\ln(x))[/math][br][math] D=0 \to r=\frac{1-b}{2} \to y_1 = x^{\frac{1-b}{2}}\:\,\:y_2 = \ln(x)\,x^{\frac{1-b}{2}}[/math][br][br]For the [math]D<0[/math] case the Euler Identity, [math]e^{i\,x}=\cos\left(x\right)+i\,\sin\left(x\right)[/math], Power rule, [math]x^r = e^{\ln(x)\,r}[/math], and real and imaginary parts must both be solutions.[br][br]The illustration below shows the two fundamental solutions where the coefficients [math]b[/math] and [math]c[/math] can be adjusted with the sliders.[br][br]