The auxiliary equation is
m2+4m+2=0m1,2=2⋅1−4±42−4⋅1⋅2m=−2+2,m=−2−2
Thus the complementary solution is
yc=(c1+xc2)e−xLet y1=e−x;y2=xe−xw=∣∣y1y1′y2y2′∣∣=∣∣e−x−e−xxe−x−xe−x+e−x∣∣=e−2x
By the method of variation of parameter particular solution is
yp=−y1∫wy2f(x)dx+y2∫wy1f(x)dxyp=x2e−x[2logx+41+logx−1]yp=4x2e−x[2logx−3]
Hence the general solution is
y(x)=(c1+xc2)e−x+4x2e−x[2logx−3]
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