# Help solving a Cauchy-Euler Equation (Differential equation help)

## Homework Statement

x2y'' + xy' + 4y = 0

y = xr
y' = r xr-1
y'' = r(r-1)xr-2

## The Attempt at a Solution

x2[r(r-1)xr-2] + x [r xr-1] + 4xr = 0

(r2-r)xr + r xr + 4xr = 0

[(r2-r) + r + 4]xr = 0

r2 - r + r + 4 = 0

r2 + 4 = 0

from the quadratic equation i know that: r = 2i and r = -2i

so y = c1x2i + c2x-2i

my question is how can i remove the imaginary number from the answer.

ideasrule
Homework Helper
x^2i can be expressed as e^(ln x^2i). Use Euler's equation to express that in terms of sine and cosine and you're done.

If the roots of your characteristic equation $r^2 + (b-1)r + c = 0$ (here a = 1, where a is the coefficient of r^2, b = 1, and c = 4) are complex (occurs since b^2 - 4ac < 0), then the general solution is

$$y= c_1 e^{\alpha x}\cos{\beta x} + c_2 e^{\alpha x}\sin{\beta x}$$

where

$$r_1 = \overline{r_2} = \alpha + i\beta = \frac{-b}{2a} + i\frac{\sqrt{4ac - b^2}}{2a}$$

so that

$$\alpha = \frac{-b}{2a} \mbox{ and } \beta = \frac{\sqrt{4ac - b^2}}{2a}$$

(All you are doing is reducing Cauchy-Euler equation to a homogeneous linear differential equation with constant coefficients)