Question

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

Answer 1

Answer:

$y(x)\ =\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.{(4x)}^{m+1}}{2m!}\ +\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.(4x){m+1}}{(2m+1)!}.$

Step-by-step explanation:

Given differential equation is

• y'' + 4y = 0           (1)

We have to find the power series solution of given differential equation about the ordinary point x = 0.

Power series solution of any given differential equation can be given by

$y(x)\ =\ C_0+C_1.x+C_2.\dfrac{x^2}{2!}+C_3.\dfrac{x^3}{3!}+....$

       $=\ \sum_{n=0}^{\infty}C_n.\dfrac{x^n}{n!}$

$=>y'(x)\ =\ \sum_{n=1}^{\infty}\dfrac{n.C_n.x^{n-1}}{n!}$

$=>y''(x)\ =\ \sum_{n=2}^{\infty}\dfracf{n.(n-1)C_n.x^{n-2}}{n!}$

Now, by putting these values in equation (1), we have

$\sum_{n=2}^{\infty}\dfracf{n.(n-1)C_n.x^{n-2}}{n!}\ +\ 4x\sum_{n=0}^{\infty}C_n.\dfrac{x^n}{n!}\ =\ 0$

$=>\ \sum_{n=0}^{\infty}\dfracf{(n+1).(n+1)C_{n+2}.x^{n}}{n!}\ +\ 4x\sum_{n=0}^{\infty}C_n.\dfrac{x^n}{n!}\ =\ 0$

$=>\ \sum_{n=0}^{\infty}[(n+1).(n+2)C_{n+2}+4xC_n]x^n\ =\ 0$

$=>\ (n+1).(n+2).C_{n+2}+4x.C_n\ =\ 0$

$=>\ C_{n+2}\ =\ \dfrac{-4x}{(n+1)(n+2)}.C_n$

for n = 0

$C_2\ =\ \dfrac{-4x}{2}.C_0$

for n = 1

$C_3\ =\ \dfrac{-4x}{6}.C_1$

for n = 2

$C_4\ =\ \dfrac{-4x}{12}.C_2$

        $=\ \dfrac{16x^2}{24}.C_0$

for n = 3

$C_5\ =\ \dfrac{-4x}{20}.C_3$

        $=\ \dfrac{16x^2}{120}$

for n=4

$C_6\ =\ \dfrac{-4x}{30}C_4$

        $=\ \dfrac{-64.x^3}{720}.C_0$

As we can see for

for even value of n i.e n = 2m where m is any integer.

$C_2m\ =\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.{(4x)}^{m+1}}{2m!}$

for odd value of n i.e n =2m+1 , where m is any integer.

$C_{2m+1}\ =\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.(4x)^{m+1}}{(2m+1)!}$

So, the power series solution about the ordinary point x=0, can be given by

$y(x)\ =\ \sum_{n=0}^{\infty}\dfrac{C_n.x^n}{n!}$

       $=\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.{(4x)}^{m+1}}{2m!}\ +\ \sum_{m=0}^{\infty}\dfrac{-(-1)^m.(4x)^{m+1}}{(2m+1)!}.$