Given: \(\displaystyle{x}{y}^{{{''}}}+{\left({2}+{2}{x}\right)}{y}^{{'}}+{2}{y}={0}\) has a solution \(\displaystyle{y}_{{{1}}}{\left({x}\right)}={x}^{{-{1}}}\),

amantantawq5l

amantantawq5l

Answered question

2022-03-26

Given: xy+(2+2x)y+2y=0 has a solution y1(x)=x1, find the general solution of the differential equation xy+(2+2x)y+2y=8e2x

Answer & Explanation

yaum3xg1

yaum3xg1

Beginner2022-03-27Added 12 answers

Step 1
Substituting yp=c1(x)x+c2(x)2xe2x into the complete ODE we obtain
2c1(x)+c1 (x)8e2x12e2x(2c2(x)c2 (x))=0
now as c1,c2 are independents, we choose those which obey
{2c1'(x)+c1''(x)8e2x=02c2'(x)c2''(x)=0
Step 2
obtaining as solutions
{c1(x)=e2x+μ1e2x+μ2c2(x)=μ3e2x+μ4
from those we choose a particular solution so we follow with μ1=μ2=μ3=μ4=0 or c1(x)=e2x, c2(x)=0 and finally we have
y=c1x+c2xe2x+1xe2x

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