Find the particular solution of the differential equation that satisfies the initial condition. Write it out as an equation

Question

Differential Equation: 20xy' - ln(x^5) = 0, x>0
Initial condition: y(1) = 13

Answer 1

20xy' - ln(x^5) = 0
y' = ln(x^5)/(20x) = 1/4 lnx/x
y = 1/8 (lnx)^2 + C
since y(1) = 13
y = 1/8 (lnx)^2 + 13

Answer 2

Thank you so much that really helps!