Given differential equation is ydy=(2+logex)dx
Integrating both sides, we get:
∫ydy=∫(2+logex)dx
logey=2x+xlogex−x+C
logey=x+xlogex+C
Since the curve passes through (1,e), substitute x=1 and y=e:
logee=1+1⋅loge1+C
1=1+0+C⇒C=0
The equation of the curve is logey=x+xlogex
To find f(e), substitute x=e:
logey=e+elogee
logey=e+e(1)=2e
y=e2e
Answer: e2e