
Bobosharif S. answered 11/11/20
PhD in Math, MS's in Calulus
y=(lnx+C)/x
y'=[(1/x)*x-(lnx+C)]/x2=(1-lnx-C)/x2
Now plug in y and y' into equation x2y' + xy = 1
x2((1-lnx-C)/x2 ) + x( lnx+C)/x = 1-lnx-C-lnx+C=1
So y=(lnx+C)/x is a general solution of the equation x2y' + xy = 1
Now
a) y(7) =6 and we find C.
6=(ln7+C)/7
42=ln7+C
C=42-ln7
So the particular solution with condition y(7)=6 is
y=(lnx+42-ln7)/x = (ln(x/7)+42)/x
b) is similar