Differential Equations, Frobenius' Method

In summary: We can rewrite the second sum as \sum_{n=0}^{\infty} (c_{n-4} - c_n (n+r)+ c_n (n+r)(n+r-1))x^{n+r-1}=0Now c_0 is not zero, so we can divide by it.Now let's use the notation a_n= c_{n
  • #1

Homework Statement



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

Homework Equations



y = Ʃ[n=0 to infinity]cnxn+r
y' = Ʃ[n=0 to infinity](n+r)cnxn+r-1
y'' = Ʃ[n=0 to infinity](r+r)(n+r-1)cnxn+r-2

The Attempt at a Solution



Plugging these values into the differential equation, I got

xr{Ʃ[n=0 to infinity](n+r)(n+r-1)cnxn-1 - Ʃ[n=0 to infinity](n+r)cnxn-1 + Ʃ[n=0 to infinity]4cnxn+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)
 
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  • #2
So you have
[tex] x^3y = \sum_{n=0}^{\infty} c_n x^{n+r+3}[/tex]
[tex] y' = \sum_{n=0}^{\infty} c_n (n+r)x^{n+r-1}[/tex]
[tex] xy'' = \sum_{n=0}^{\infty} c_n (n+r)(n+r-1)x^{n+r-1}[/tex]

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
[tex] x^3y = \sum_{m=0}^{\infty} c_m x^{m+r+3}[/tex]
then let m=n-4, which when m=0 gives n=4, subtituting in gives
[tex] x^3y =\sum_{n=4}^{\infty} c_{n-4} x^{n+r-1} [/tex]


Putting this all together we get
[tex] \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[/tex]

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

Now the indicial equation is given by the lowest power of x, this occurs when n=0
 
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What is Frobenius' method?

Frobenius' method is a technique used to find solutions to linear differential equations with variable coefficients. It allows us to find solutions in the form of power series, which can be useful in solving certain types of differential equations.

When is Frobenius' method used?

Frobenius' method is typically used when the coefficients in a linear differential equation are not constant, making it difficult to solve using other methods such as separation of variables or substitution.

What are the steps involved in using Frobenius' method?

The first step is to rewrite the differential equation in standard form, with the highest derivative term isolated. Then, we assume a solution in the form of a power series and substitute it into the equation. This leads to a recurrence relation, which can be used to find the coefficients of the power series. Finally, the solution is obtained by substituting the coefficients into the power series.

What are some limitations of Frobenius' method?

Frobenius' method can only be used to solve linear differential equations with variable coefficients. Additionally, it may not always be possible to find a solution in the form of a power series, in which case this method cannot be applied.

What are some applications of Frobenius' method?

Frobenius' method is commonly used in physics and engineering to solve differential equations that arise in areas such as heat transfer, fluid dynamics, and quantum mechanics. It is also used in mathematical modeling to understand and predict natural phenomena.

Suggested for: Differential Equations, Frobenius' Method

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