Consider the DEd2ydx2+2dydx+y=xwhich is linear with constant coefficients.First we will work on solving the corresponding homogeneous equation. The auxiliary equation (using m as your variable) is m2+2m+1 =0 which has root 1.Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: ex,xexAnswer PreviewYou Enterede^(-x), x*[e^(-x)]Preview of Your Answerex,xex to do reduction of order.y2=uex.Then (using the prime notation for the derivatives)y2= uex+uexy2= uex+uex+uexuexSo, plugging y2 into the left side of the differential equation, and reducing, we gety2+2y2+1y2=uexSo now our equation is exu=x. To solve for u we need only integrate xe1x twice,