Question

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

Answer 1

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

Explanation:

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}`

Answer 2

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

Explanation:

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)`