Differential Equations, Frobenius' Method

[jakobsandberg]

Homework Statement

Find the indicial roots of the following Differential Equation: xy'' - y' + x³y = 0

Homework Equations

y = Ʃ[n=0 to infinity]c_nx^n+r
y' = Ʃ[n=0 to infinity](n+r)c_nx^n+r-1
y'' = Ʃ[n=0 to infinity](r+r)(n+r-1)c_nx^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_nx^n-1 - Ʃ[n=0 to infinity](n+r)c_nx^n-1 + Ʃ[n=0 to infinity]4c_nxⁿ⁺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 x³ as their first term [the first two sums starting from n=4]. However, this left me with the following equation:

r(r-1)c₀x^-1 - rc₀x^-1 + r(r+1)c₁ - (r+1)c₁ + (r+1)(r+2)c₂x - (r+2)c₂x - (r+2)(r+3)c₃x² - (r+3)c₃x² + [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=c₁cos(x²) + c₂sin(x²)

 

[lanedance]

So you have
x^3y = \sum_{n=0}^{\infty} c_n x^{n+r+3}
y' = \sum_{n=0}^{\infty} c_n (n+r)x^{n+r-1}
xy'' = \sum_{n=0}^{\infty} c_n (n+r)(n+r-1)x^{n+r-1}

so let's shift the equations, so they have the same form for powers of x, as follows:

lets leave the 2nd & 3rd leave unchanged, but replace the 1st so the terms match, first replace n with m
x^3y = \sum_{m=0}^{\infty} c_m x^{m+r+3}
then let m=n-4, which when m=0 gives n=4, subtituting in gives
x^3y =\sum_{n=4}^{\infty} c_{n-4} x^{n+r-1}


Putting this all together we get
\sum_{n=4}^{\infty} c_{n-4} x^{n+r-1}- \sum_{n=0}^{\infty} c_n (n+r)x^{n+r-1}+ \sum_{n=0}^{\infty} c_n (n+r)(n+r-1)x^{n+r-1}=0

Now let's group terms
\sum_{n=0}^{3} c_n(- (n+r)+ (n+r)(n+r-1))x^{n+r-1}+ <br /> \sum_{n=4}^{\infty} (c_{n-4} - c_n (n+r)+ c_n (n+r)(n+r-1))x^{n+r-1}=0

Now the indicial equation is given by the lowest power of x, this occurs when n=0