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`        Please answer this question::(e^x +1)ydy - (y2+1)e^xdx = 0  Given x=0 y=0`
8 months 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.
```
8 months ago
```							Please sir do the full solutionI have done 70% of the problem but can't go further please help.......
```
8 months ago
```							(ex + 1)ydy – (y2 + 1)exdx = 0 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 = dtand similarly ex + 1 = u, then ex dx = du 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 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.]
```
7 months 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
```
7 months ago
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