Question

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

Answer 1

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

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.(x^2)/(2!)+C_3.(x^3)/(3!)+....`

`=\ \sum_(n=0)^(\infty)C_n.(x^n)/(n!)`

`=>y'(x)\ =\ \sum_(n=1)^(\infty)(n.C_n.x^(n-1))/(n!)`

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

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

`\sum_(n=2)^(\infty)\fracf{n.(n-1)C_n.x^(n-2)}{n!}\ +\ 4x\sum_(n=0)^(\infty)C_n.(x^n)/(n!)\ =\ 0`

`=>\ \sum_(n=0)^(\infty)\fracf{(n+1).(n+1)C_(n+2).x^(n)}{n!}\ +\ 4x\sum_(n=0)^(\infty)C_n.(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)\ =\ (-4x)/((n+1)(n+2)).C_n`

for n = 0

`C_2\ =\ (-4x)/(2).C_0`

for n = 1

`C_3\ =\ (-4x)/(6).C_1`

for n = 2

`C_4\ =\ (-4x)/(12).C_2`

`=\ (16x^2)/(24).C_0`

for n = 3

`C_5\ =\ (-4x)/(20).C_3`

`=\ (16x^2)/(120)`

for n=4

`C_6\ =\ (-4x)/(30)C_4`

`=\ (-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)\frac{-(-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)(-(-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)(C_n.x^n)/(n!)`

`=\ \sum_(m=0)^(\infty)\frac{-(-1)^m.{(4x)}^(m+1)}{2m!}\ +\ \sum_(m=0)^(\infty)(-(-1)^m.(4x)^(m+1))/((2m+1)!).`