Power Series Solutions of Differential Equations
Key Questions

Assuming you know how to find a power series solution for a linear differential equation around the point
#x_0# , you just have to expand the source term into a Taylor series around#x_0# and proceed as usual.This may add considerable effort to the solution and if the power series solution can be identified as an elementary function, it's generally easier to just solve the homogeneous equation and use either the method of undetermined coefficients or the method of variation of parameters.

The solution is
#y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x# ,where
#c_0# is any constant.Let us look at some details.
Let
#y=sum_{n=0}^infty c_n x^n#
#y'=sum_{n=1}^infty nc_n x^{n1}=sum_{n=0}^infty(n+1)c_{n+1}x^n# So, we can rewrite
#y'y=0# as#sum_{n=0}^infty (n+1)c_{n+1} x^nsum_{n=0}^infty c_n x^n=0# by combining the summations,
#Rightarrow sum_{n=0}^infty[(n+1)c_{n+1}c_n]x^n=0# so, we have
#(n+1)c_{n+1}c_n=0 Rightarrow c_{n+1}=1/{n+1}c_n# Let us observe the first few terms.
#c_1=1/1c_0=1/{1!}c_0# #c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0# #c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0#
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#c_n=1/{n!}c_0# Hence, the solution is
#y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x# ,where
#c_0# is any constant. 
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^(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


Let us solve the differential equation
#y''=y# by Power Series Method.Let
#y=sum_{n=0}^inftyc_n x^n# , where#c_n# is to be determined.
By taking derivatives term by term,
#y'=sum_{n=1}^{infty}nc_nx^{n1}#
and
#y''=sum_{n=2}^infty n(n1)c_nx^{n2}# So,
#y''=y# becomes
#sum_{n=2}^infty n(n1)c_nx^{n2}=sum_{n=0}^inftyc_n x^n#
by shifting the indices on the summation on the left by 2,
#sum_{n=0}^infty(n+2)(n+1)c_{n+2}x^n=sum_{n=0}^inftyc_n x^n# By matching each coefficients,
#(n+2)(n+1)c_{n+2}=c_n Rightarrow c_{n+2}=c_n/{(n+2)(n+1)}# Let us observe the first few even terms,
#c_2=1/{2cdot1}c_0=1/{2!}c_0#
#c_4=1/{4cdot3}c_2=1/{4cdot3}cdot1/{2!}c_0=1/{4!}c_0#
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#c_{2n}=1/{(2n)!}c_0# Let us observe the first few odd terms,
#c_3=1/{3cdot2}c_1=1/{3!}c_1#
#c_5=1/{5cdot4}c_3=1/{5cdot4}cdot1/{3!}c_1=1/{5!}c_1#
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#c_{2n+1}=1/{(2n+1)!}c_1# Now, we can find the solution
#y# .
#y=sum_{n=0}^infty c_nx^n#
by separating even terms and odd terms,
#=sum_{n=0}^inftyc_{2n}x^{2n}+sum_{n=0}^inftyc_{2n+1}x^{2n+1}#
by using the formulas for#c_{2n}# and#c_{2n+1}# above,
#=c_0sum_{n=0}^inftyx^{2n}/{(2n)!}+c_1sum_{n=0}^infty x^{2n+1}/{(2n+1)!}# Recall:
#coshx=sum_{n=0}^infty x^{2n}/{(2n)!}#
#sinhx=sum_{n=0}^infty x^{2n+1}/{(2n+1)!}# Hence,
#y=c_0coshx+c_1sinhx# , where#c_0# and#c_1# are any constants.
Questions
Power Series

Introduction to Power Series

Differentiating and Integrating Power Series

Constructing a Taylor Series

Constructing a Maclaurin Series

Lagrange Form of the Remainder Term in a Taylor Series

Determining the Radius and Interval of Convergence for a Power Series

Applications of Power Series

Power Series Representations of Functions

Power Series and Exact Values of Numerical Series

Power Series and Estimation of Integrals

Power Series and Limits

Product of Power Series

Binomial Series

Power Series Solutions of Differential Equations