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

1 Answer
Jul 23, 2018

Answer:

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^oo a_n x^n#

  • #implies y^' = sum_1^oo n a_n x^(n-1) qquad qquad y^('')= sum_2^oo n (n-1)a_n x^(n-2)#

With this power series:

#y''+2xy'+y = 0#

#implies underbrace(sum_2^oo n (n-1)a_n x^(n-2))_(=sum_0^oo (m+2) (m+1)a_(m+2) x^(m)) + 2x sum_1^oo n a_n x^(n-1) + sum_0^oo a_n x^n= 0#

#implies sum_0^oo (n+2) (n+1)a_(n+2) x^n + 2 sum_1^oo n a_n x^n + sum_0^oo a_n x^n= 0#

#implies 2 a_2 + sum_1^oo (n+2) (n+1)a_(n+2) x^n + 2 sum_1^oo n a_n x^n + a_0 + sum_1^oo a_n x^n= 0#

#:. underbrace(2 a_2 + a_0)_(=0) + sum_1^oo ( underbrace( (n+2) (n+1)a_(n+2) + (2 n +1)a_n )_(=0) )x^n = 0#

This insists that the coefficient of the #x^0# term is zero, but also that the coefficient of every #x^(i gt 0)# term is also zero, as the homogeneous DE requires.

The recurrence relation is:

  • #a_(n+2) = - ((2 n +1))/((n+2) (n+1)) 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\ y_O + c_2 \ 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:

Socratic

I'd recommend: this