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## Homework Statement

Find the indicial roots of the following Differential Equation: xy'' - y' + x

^{3}y = 0

## Homework Equations

y = Ʃ[n=0 to infinity]c

_{n}x

^{n+r}

y' = Ʃ[n=0 to infinity](n+r)c

_{n}x

^{n+r-1}

y'' = Ʃ[n=0 to infinity](r+r)(n+r-1)c

_{n}x

^{n+r-2}

## The Attempt at a Solution

Plugging these values into the differential equation, I got

x

^{r}{Ʃ[n=0 to infinity](n+r)(n+r-1)c

_{n}x

^{n-1}- Ʃ[n=0 to infinity](n+r)c

_{n}x

^{n-1}+ Ʃ[n=0 to infinity]4c

_{n}x

^{n+3}} = 0

The three sums must produce the x to the same exponent, so I tried pulling out the first 4 terms of the first two sums, so the three sums would each output x

^{3}as their first term [the first two sums starting from n=4]. However, this left me with the following equation:

r(r-1)c

_{0}x

^{-1}- rc

_{0}x

^{-1}+ r(r+1)c

_{1}- (r+1)c

_{1}+ (r+1)(r+2)c

_{2}x - (r+2)c

_{2}x - (r+2)(r+3)c

_{3}x

^{2}- (r+3)c

_{3}x

^{2}+ [remaining sums] = 0.

How do I solve for r with this equation? I don't know how to find the roots.

[the solution to the DE is y=c

_{1}cos(x

^{2}) + c

_{2}sin(x

^{2})

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