If the curve y=f(x) passes through the point (1, e) and satisfies the differential equation dy=y (2+ e x )dx, x>0, then f(e) is equal to :
- ee
- ee2
- e2e
- e2e
Answer
(3) e2e
Solution
dyy=(2+ ex) dx
Integrating: ey=2x+(x ex-x)+C=x+x ex+C
At (1,e): 1=1+0+C C=0
ey=x(1+ ex)
At x=e: ey=e(1+1)=2e
∴ f(e)=e2e
Answer: (3)