What is a solution to the differential equation #y' + y + e^(x) x^(4) = 0#?

quick re-write, this is not separable so we put it in form for Integrating Factor (I.F.)

#y' + y =- e^(x) x^(4) #

I.F. is #exp (int dx) = e^x#

#e^x y' + e^x y =- e^(2x) x^(4) #

#(e^x y)' =- e^(2x) x^(4) #

#e^x y =- int \ e^(2x) x^(4) \ dx qquad star#

#int \ e^(2x) x^(4) \ dx# requires 4 rounds of IBP, basically reducing that #x^4# term down, meaning

#int \ e^(2x) x^(4) \ dx = 1/4 e^(2 x) (2 x^4-4 x^3+6 x^2-6 x+3)+ C#

so #star# becomes

#e^x y =-1/4 e^(2 x) (2 x^4-4 x^3+6 x^2-6 x+3)+ C#

# y =-1/4 e^( x) (2 x^4-4 x^3+6 x^2-6 x+3)+ C e^{-x}#

The differential equation is linear non-homogeneus.

We propose a solution with structure

#y = e^x p(x)#

with #p(x) = sum_{k=0}^n a_kx^k# a polynomial, and substituting into the equation, we have

#e^xp(x)+e^xp'(x)+e^xp(x)+e^x x^4 = 0#

but #e^x ne 0# so

#2p(x)+p'(x)+x^4=0#

or

#2sum_{k=0}^n a_kx^k+sum_{k=0}^n ka_kx^{k-1}+x^4=0#

Choosing #n = 4# and equating same power coefficients

#{(2 a_0 + a_1=0), (2 a_1+ 2 a_2=0), (2 a_2 + 3 a_3=0), (2 a_3 + 4 a_4=0), ( 1 + 2 a_4=0) :}#

solving we have

#{a_0= -3/4, a_1= 3/2, a_2= -3/2, a_3= 1, a_4= -1/2}#

so the solution is

#y = e^x/4 (-3 + 6 x - 6 x^2 +4 x^3 - 2x^4+C_0)#