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

Jul 11, 2016

$y = - \frac{1}{4} {e}^{x} \left(2 {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 6 x + 3\right) + C {e}^{- x}$

#### 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 $\exp \left(\int \mathrm{dx}\right) = {e}^{x}$

${e}^{x} y ' + {e}^{x} y = - {e}^{2 x} {x}^{4}$

$\left({e}^{x} y\right) ' = - {e}^{2 x} {x}^{4}$

${e}^{x} y = - \int \setminus {e}^{2 x} {x}^{4} \setminus \mathrm{dx} q \quad \star$

$\int \setminus {e}^{2 x} {x}^{4} \setminus \mathrm{dx}$ requires 4 rounds of IBP, basically reducing that ${x}^{4}$ term down, meaning

$\int \setminus {e}^{2 x} {x}^{4} \setminus \mathrm{dx} = \frac{1}{4} {e}^{2 x} \left(2 {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 6 x + 3\right) + C$

so $\star$ becomes

${e}^{x} y = - \frac{1}{4} {e}^{2 x} \left(2 {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 6 x + 3\right) + C$

$y = - \frac{1}{4} {e}^{x} \left(2 {x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 6 x + 3\right) + C {e}^{- x}$

Jul 11, 2016

$y = {e}^{x} / 4 \left(- 3 + 6 x - 6 {x}^{2} + 4 {x}^{3} - 2 {x}^{4} + {C}_{0}\right)$

#### Explanation:

The differential equation is linear non-homogeneus.

We propose a solution with structure

$y = {e}^{x} p \left(x\right)$

with $p \left(x\right) = {\sum}_{k = 0}^{n} {a}_{k} {x}^{k}$ a polynomial, and substituting into the equation, we have

${e}^{x} p \left(x\right) + {e}^{x} p ' \left(x\right) + {e}^{x} p \left(x\right) + {e}^{x} {x}^{4} = 0$

but ${e}^{x} \ne 0$ so

$2 p \left(x\right) + p ' \left(x\right) + {x}^{4} = 0$

or

$2 {\sum}_{k = 0}^{n} {a}_{k} {x}^{k} + {\sum}_{k = 0}^{n} k {a}_{k} {x}^{k - 1} + {x}^{4} = 0$

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

$\left\{{a}_{0} = - \frac{3}{4} , {a}_{1} = \frac{3}{2} , {a}_{2} = - \frac{3}{2} , {a}_{3} = 1 , {a}_{4} = - \frac{1}{2}\right\}$

so the solution is

$y = {e}^{x} / 4 \left(- 3 + 6 x - 6 {x}^{2} + 4 {x}^{3} - 2 {x}^{4} + {C}_{0}\right)$