# Please answer this question::(e^x +1)ydy - (y2+1)e^xdx = 0  Given x=0 y=0

Grade:11

## 4 Answers

Arun
25757 Points
4 years ago
Dear student

It can be written as

y dy/(y^2 +1) = e^x dx /(e^x + 1)

Now I think you can do.

If you find any difficulty, please feel free to ask me again.
Aniket debnath
14 Points
4 years ago
Please sir do the full solution
I have done 70% of the problem but can't go further please help.......
Samyak Jain
333 Points
3 years ago
(ex + 1)ydy – (y2 + 1)exdx = 0
$\Rightarrow$ ydy/(y2 + 1) = ex dx /(ex + 1)
2ydy/(y2 + 1) = 2ex dx /(ex + 1)
Now if  y2 + 1 = t, then differentiating both sides we get 2ydy = dt
and similarly ex + 1 = u, then ex dx = du
$\dpi{80} \therefore$ dt / t = 2 du / u .  Integratin both sides, we get
ln(t)  =  2 ln(u)  +  lnc , where I’ve taken lnc as constant of integration and ln(t) means loge(t).
ln(t) – 2 ln(u)  =  lnc  i.e. ln(t) – ln(u)2  =  lnc  or  ln(t/u2) = lnc
$\dpi{80} \Rightarrow$ t/u2 = c             [using rules of logarithm]
Substitute t as y2 + 1 and u as ex + 1 to get
(y2 + 1)/(ex + 1)2 = c      [You can rearrange the result to get the desired answer.]
Samyak Jain
333 Points
3 years ago
Apply the condition : x = 0, y = 0 to get constant c
(y2 + 1)/(ex + 1)2 = c
(02 + 1) / (e0 + 1)2 = c  =  1/(2)2 = 1/4
(y2 + 1) / (ex + 1)2  = 1/4
i.e.   4(y2 + 1)  =  (ex + 1)2