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Mathematics Question on Differential equations

The solution of differential equation ydxdy=x(logex−logey+1), x>0,y>0y\frac {dx}{dy}=x(log_e x-log_e y+1),\ x>0, y>0 and passing through (e,1)(e,1) is

A

∣loge(yx)∣=y2|log_e(\frac yx)| = y^2

B

2∣loge(xy)∣=y2|log_e(\frac xy)| = y

C

∣loge(yx)∣=x|log_e(\frac yx)| = x

D

∣loge(xy)∣=y|log_e(\frac xy)| = y

Answer

∣loge(xy)∣=y|log_e(\frac xy)| = y

Explanation

Solution

The correct option is (D): ∣loge(xy)∣=y|log_e(\frac xy)| = y