Question 1184077
Let {{{u = y^(1-4) = y^(-3)}}}.  Substituting this into the original D.E. and simplifying, you get


{{{"u'" + (21/x)u = -3/x^5}}}.


The integrating factor is {{{e^(21int(dx/x)) = x^21}}}.

After multiplying D.E. with the integrating factor and integrating, 

you get 


{{{ux^21 = (-3/17)x^17 + C}}}.  ===> {{{y^(-3)x^21 = (-3/17)x^17 + C}}}, or  {{{1/y^3 = -3/(17x^4) + C/x^21}}}.


solving for C when y(1) = 1 gives {{{C = 20/17}}}, so that


{{{1/y^3 = -3/(17x^4) + 20/(17x^21)}}}.