Solving DE Auxilary Equation: x^{2}y" - xy' + y = 0

[Lancelot59]

I need to find a solution to:
$$x^{2}y"-xy'+y=0$$ in the form of $$y=x^{r}$$ where r is a constant.

I started by finding the appropriate derivatives:
$$y=x^{r}$$
$$y'=rx^{r-1}$$
$$y"=r^{2}x^{r-2}$$

Then substituting in:
$$x^{2}(r^{2}x^{r-2})-x(rx^{r-1})+x^{r}=0$$
which simplifies to:
$$r^{2}-r+1=0$$

I then solved and got the complex roots:
$$\frac{1\pm i\sqrt{3}}{2}$$

I'm not sure what to do next. The examples I've seen so far have separated out the imaginary part using identities, where the function is exponential.

 

[lippyka]

However, this is not the case here.Any help is appreciated. Thank you!The solution is r=(1 ± i√3)/2. This means that the solution to your equation can be written as:y=x^(1 ± i√3)/2.