value

Is the answer to the question is 3 ??

is the answer approx. = 2

dy/dx = y + 1.

integrating,

y = yx + x + c, where c is constant.

putting x = 0 we get,

y = c, but y =1 at x=0 (given)

c=1.

therefore, y = yx + x + 1.
=> y = (x + 1)/(1 - x).
at x= ln2

y = (ln2 + 1)/(1 - ln2)

how could u integrate this way??

sorry i was damn wrong there....
i didn't pay heed to that y..

dy/dx = y+1.
dy/(y+1) = dx.

integrating,
=> ∫dy/(y+1) =∫dx.
=> ln(y+1) = x + c.

at x=0 and y=1, we get,

c = ln2
thus ln(y+1) = x + ln2
therefore, at x=ln2..

ln(y+1) = 2ln2 = ln4.
therefore y=3.