Question

Find a power series solution of the differential equation y" + 4xy = 0 about the ordinary point x = 0.

Answer 1

Answer:

$y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])$

Step-by-step explanation:

$y"+4xy=0$

Assume that the problem has a solution of the form:

$y=f(x)=\sum_0^{\infty}(c_nx^n)$

then calculate the derivative:

$y'=\sum_1^{\infty}(nc_nx^{n-1})$

and the second derivative:

$y"=\sum_2^{\infty}(n(n-1)c_nx^{n-2})$

In these sums the subindex indicates the first non cero term.

Making the substitution in the original equation:

$\sum_2^{\infty}(n(n-1)c_nx^{n-2})+4x\sum_0^{\infty}(c_nx^{n})=0$

Now look at the subindex of the sum: they do not match; you need to make them match:

$\sum_0^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_0^{\infty}(c_nx^{n+1})=0$

Now they match and they could be joined together on the same sum symbol, however the x term doesn't have the same exponent (after multiplying the x from the original formula with the one inside the sum). Therefore you need to expand the second derivative term and displace the other one (as we did in the previous step) to match them both:

$2c_2+\sum_1^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_1^{\infty}(c_{n-1}x^{n})=0$

Now you can join them together:

$2c_2+\sum_1^{\infty}[(n+2)(n+1)c_{n+2}x^{n}+4(c_{n-1}x^{n})]=0$

$2c_2+\sum_1^{\infty}x^n=0$

If the solution is valid for all x, then all terms must be independantly 0. therefore:

$a_2=0\\(n+2)(n+1)c_{n+2}+4(c_{n-1})=0\\c_{n+2}=\frac{-4(c_{n-1})}{(n+2)(n+1)}$

Now you need to use that recurrent function to calculate the coefficients:

$n=1 \rightarrow c_3=\frac{-4c_0}{3*2}=-\frac{2}{3} c_0\\n=2 \rightarrow c_4=\frac{-4c_1}{4*3}=-\frac{1}{3} c_1\\n=3 \rightarrow c_5=\frac{-4c_2}{5*4}=-\frac{1}{5} c_2=0\\n=4 \rightarrow c_6=\frac{-4c_3}{6*5}=-\frac{2}{15} c_3\\n=5 \rightarrow c_7=\frac{-4c_4}{7*6}=-\frac{2}{21} c_4\\n=6 \rightarrow c_8=\frac{-4c_5}{8*7}=0\\n=7 \rightarrow c_9=\frac{-4c_6}{9*8}=-\frac{1}{18} c_6$

you found that for each n divisible by 3, therefore for each $c_{3n-1}$ the coefficient is 0:

$c_2=c_5=c_8....=c_{3n-1}=0$

now look at the terms $a_3,a_6,a_9$, they are all recurrent to c_0, therefore you can wirte a rule for them:

$c_{3n} = \frac{(-4)^nc_0}{(2*3)(5*6)...(3n-1)(3n)}$

Finally, look at the terms $a_4,a_7$, they recurr to c_1, therefore:

$c_{3n+1} = \frac{(-4)^nc_1}{(3*4)(6*7)...(3n)(3n+1)}$

So you have two constants that are independant, c_0 and c_1.

Therefore the solution must be writen in terms of these two arbitrary constants:

$y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])$