How do you use Power Series to solve the differential equation #y''+2xy'+y=0# ?
See below
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^(n1) qquad qquad y^('')= sum_2^oo n (n1)a_n x^(n2)# With this power series:
#y''+2xy'+y = 0# #implies underbrace(sum_2^oo n (n1)a_n x^(n2))_(=sum_0^oo (m+2) (m+1)a_(m+2) x^(m)) + 2x sum_1^oo n a_n x^(n1) + 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 superimposed 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:
I'd recommend: this
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To solve the differential equation ( y'' + 2xy' + y = 0 ) using power series, we assume that the solution can be represented by a power series:
[ y(x) = \sum_{n=0}^\infty a_nx^n ]
Substituting this series into the given differential equation and equating coefficients of like powers of ( x ) yields a recursion relation for the coefficients ( a_n ). By solving this recursion relation, we can find the coefficients of the power series solution.
First, differentiate ( y(x) ) twice and substitute into the differential equation. Then, equate coefficients of like powers of ( x ) to obtain the recursion relation. Solve this relation to find the coefficients ( a_n ).
Finally, the power series solution ( y(x) ) is the sum of the terms with the found coefficients ( a_n ), and it converges within a certain radius of convergence determined by the solution's behavior.
This method allows us to find a series representation for the solution of the given differential equation.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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