Let us use the transformation x=et. Then
yx′=yt′e−t, yx2′′=(yt2′′−yt′)e−2t.
We get the equation for the function y(t)
e2t(y′′−y′)e−2t−ety′e−t+y=t, which is equivalent to
y′′−2y′+y=t
Corresponding homogeneous differential equation
y′′−2y′+y=0 Characteristic (auxiliary) equation
r2−2r+1=0
r1=r2=1 The general solution of the homogeneous differential equation is
yh=c1et+c2tet
Find the particular solution of the non homogeneous differential equation
yp=At+B
yp′=A
yp′′=0 Substitute
0−2A+At+B=t
A=1,B=2 The general solution of the non homogeneous differential equation is
y=c1et+c2tet+t+2
y(x)=c1x+c2xlnx+lnx+2
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