40xy' -ln(x^8) = 0
Initial Condition: y(1)=31
Work:
40x (dy/dx) - ln(x^8) = 0
40x(dy/dx) = ln (x^8)
dy/dx = ln(x^8) / 40x
Sdy = Sln(x^8) / 40x
And then...stuck (if its even right thus far)
Because is $\displaystyle \ln x^{8} = 8\ \ln x$ the DE becomes...
$\displaystyle \displaystyle y^{'} = \frac{\ln x}{5 x}$ (1)
... where the variables are separated so that the solution is immediate...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$