# Second Order Differential Equation - Power Series Solution

Hi there, I have a question regarding Differential Equations which can be located here http://none-questions.blogspot.com/2011/09/d-q7.html. can someone please take a look? Thanks.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

We wish to solve the following differential equation using power series:

xy'' - 2y + 9x^5 y = 0.

First write out y(x) as a power series, i.e.

y(x) = sum_{k=0}^infty(c_k x^k).

Then we have

y'(x) = sum_{k=1}^infty[k c_k x^(k-1)]

and

y''(x) = sum_{k=2}^infty[k(k-1) c_k x^(k-2)]

whence

0 = xy'' - 2y + 9x^5 y

= sum_{k=2}^infty[k(k-1) c_k x^(k-1)] - 2 sum_{k=1}^infty[k c_k x^(k-1)]

+ 9 sum_{k'=0}^infty[c_k' x^(k'+5)].

Now let k' = k-6. Then we have

0 = sum_{k=1}^infty{[k(k-3) c_k + 9 c_(k-6)] x^(k-1)}.

Since the sum is zero, each coefficient multiplying x^(k-1) must be zero, i.e.

k(k-3) c_k + 9 c_(k-6) = 0 for all k.

Thus we have

c_k = -9 c_(k-6) / [k(k-3)]

for all k.

Assume c_0 is nonzero. Then we have

c_(6k) = -9 c_(6k-6) / [(6k)(6k-3)]

= -c_(6k-6) / [(2k)(2k-1)]

= +c_(6k-12) / [(2k)(2k-1)(2k-2)(2k-3)]

...

= (-1)^k c_0/(2k)!

Thus we have the particular solution

y_1(x) = sum_{k=0}^infty(c_k x^k) = c_0 sum_{k=0}^infty[(-1)^k x^(6k)/(2k)!] = c_0 cos(x^3).

To find a linearly independent solution, assume c_3 is nonzero as proceed as before. We have

c_(6k+3) = -9 c_(6k-3) / [(6k+3)(6k)]

= -c_(6k-3) / [(2k+1)(2k)]

= +c_(6k-9) / [(2k+1)(2k)(2k-1)(2k-2)]

...

= (-1)^k c_3/(2k+1)!

Thus we have the linearly independent solution

y_2(x) = sum_{k=0}^infty(c_k x^k) = c_3 sum_{k=0}^infty[(-1)^k x^(6k+3)/(2k+1)!] = c_3 sin(x^3).

Thus the general solution of the differential equation is

y(x) = A cos(x^3) + B sin(x^3)

where A and B are arbitrary constants.

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