Solve the initial value problem
xy′+(1+1ln(x))y=0\displaystyle {x}{y}'+{\left({1}+\frac{{1}}{{\ln{{\left({x}\right)}}}}\right)}{y}={0}xy′+(1+ln(x)1)y=0 with initial condition y(e)=1\displaystyle {y}{\left({e}\right)}={1}y(e)=1
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