# How do you use Power Series to solve the differential equation y''+2xy'+y=0 ?

Jul 23, 2018

See below

#### Explanation:

Assuming a power series solution like this:

• $y = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + {a}_{3} {x}^{3} + \ldots = {\sum}_{0}^{\infty} {a}_{n} {x}^{n}$

• $\implies {y}^{'} = {\sum}_{1}^{\infty} n {a}_{n} {x}^{n - 1} q \quad q \quad {y}^{' '} = {\sum}_{2}^{\infty} n \left(n - 1\right) {a}_{n} {x}^{n - 2}$

With this power series:

$y ' ' + 2 x y ' + y = 0$

$\implies {\underbrace{{\sum}_{2}^{\infty} n \left(n - 1\right) {a}_{n} {x}^{n - 2}}}_{= {\sum}_{0}^{\infty} \left(m + 2\right) \left(m + 1\right) {a}_{m + 2} {x}^{m}} + 2 x {\sum}_{1}^{\infty} n {a}_{n} {x}^{n - 1} + {\sum}_{0}^{\infty} {a}_{n} {x}^{n} = 0$

$\implies {\sum}_{0}^{\infty} \left(n + 2\right) \left(n + 1\right) {a}_{n + 2} {x}^{n} + 2 {\sum}_{1}^{\infty} n {a}_{n} {x}^{n} + {\sum}_{0}^{\infty} {a}_{n} {x}^{n} = 0$

$\implies 2 {a}_{2} + {\sum}_{1}^{\infty} \left(n + 2\right) \left(n + 1\right) {a}_{n + 2} {x}^{n} + 2 {\sum}_{1}^{\infty} n {a}_{n} {x}^{n} + {a}_{0} + {\sum}_{1}^{\infty} {a}_{n} {x}^{n} = 0$

$\therefore {\underbrace{2 {a}_{2} + {a}_{0}}}_{= 0} + {\sum}_{1}^{\infty} \left({\underbrace{\left(n + 2\right) \left(n + 1\right) {a}_{n + 2} + \left(2 n + 1\right) {a}_{n}}}_{= 0}\right) {x}^{n} = 0$

This insists that the coefficient of the ${x}^{0}$ term is zero, but also that the coefficient of every ${x}^{i > 0}$ term is also zero, as the homogeneous DE requires.

The recurrence relation is:

• ${a}_{n + 2} = - \frac{\left(2 n + 1\right)}{\left(n + 2\right) \left(n + 1\right)} {a}_{n}$

This suggest 2 independent solutions to the DE, one for odd terms and one for even. These can be linearly super-imposed to reach a general null solution.

Even terms:

Setting: ${a}_{0} = 1 , {a}_{1} = 0$:

{( a_0 = 1),(a_2 = - 1/2 ),(a_4 = - 5/(4*3)* - 1/2 = 5/(4!) ),(a_6 = -(9)/(6*5) * 5/(4!) = - (5*9)/(6!) ),(a_8 = -(13)/(8*7) * - (5*9)/(6!) = (5*9*13)/(8!) ):}

 implies underbrace(1/(1!) x^0)_(k = 0) + underbrace((-1)^(1) 1/(2!) x^2)\_(k = 1) + (-1)^(2) 5/(4!) x^4 + (-1)^(3) (5*9)/(6!) x^8 + (-1)^(4) (5*9*13)/(8!) x^8 + cdots + underbrace((-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k))_("but not for first 2 terms") + cdots = 0

An even solution is therefore:

y_E = 1 - x^2/2 + sum_2^oo (-1)^(k) (5* 9 * ... * (4 k - 3))/((2k)!) x^(2 k)

Odd terms:

Setting: ${a}_{0} = 0 , {a}_{1} = 1$, and copying the broad pattern:

{( a_1 = 1),(a_3 =- 3/(2*3) = - 3/(3!) ),(a_5 = - 7/(5*4)* - 3/(3!) = (3*7)/(5!) ),(a_7 = - 11/(7*6) * (3*7)/(5!) =- (3*7*11)/(7!)),(a_9 = -(3*7*11)/(7!)* -(15)/((9*8) )= (3*7*11*15)/(9!) ):}

 implies underbrace( x )_(k = 0) + underbrace((-1)^(1) 3/(3!) x^3)\_(k = 1) + (-1)^(2) (3*7)/(5!) x^5 + (-1)^(3) (3*7*11)/(7!) x^7 + cdots + underbrace((-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1)))_("but not for first term") + cdots = 0

An odd solution is therefore:

y_O =x + sum_1^oo (-1)^(k) (3* 7 * ... * (4 k -1))/((2k+1)!) x^((2 k+1))

Recognising the linearity:

$y = {c}_{1} \setminus {y}_{O} + {c}_{2} \setminus {y}_{E}$

So you have to add all that up

Finally , a screen grab from Socratic that always puts me of answering qu's like this in proper fashion:

I'd recommend: this