Question

differential equations e^xdx-(1+e^-x)^2dy=0

Answer 1

It looks separable!

Answer 2

can you help me with it? please

Answer 3

Switch the term with x, dx to the right side!

Answer 4

Can you do it?

Answer 5

Just isolate dy
Then integrate both side to get y !

Answer 6

i am trying

Answer 7

i dont get how i should integrate the right side

Answer 8

e^x dx-(1+e^-x)^2dy=0
(1+e^-x)^2dy = e^xdx
=> dy = e^xdx / (1+e^-x)^2

Answer 9

Agree?

Answer 10

yes, and what then? do we do it by partial? :/ i am so lost i followed u until there

Answer 11

Since you want y!
-> Int dy = int e^xdx / (1+e^-x)^2

Agree?

Answer 12

oh we take ln on both sides?

Answer 13

Integrate dy => y ( not ln)

Answer 14

yes this is where i am getting stuck at i can seem to find away to integrate e^x/(1-E^-x)^2 dx

Answer 15

i cant*

Answer 16

a way*

Answer 17

should i open the square?

Answer 18

Let u = 1 + e^-x
-> du = ?

Answer 19

e^-x?

Answer 20

-e^-x*

Answer 21

du = - e^-x dx

Answer 22

Is it okie for you to continue?

Answer 23

but in numerator its e^x not e^-x ? how am i suppose to substiute du?

Answer 24

=> -du = e^-x dx

Answer 25

Oh, let me think!

Answer 26

yes but still on top it is e^x and my (-)du is e^-x? isnt du suppose to by e^+x dx?

Answer 27

in order to substitute

Answer 28

I'm thinking of - 1/ du = e^x dx

Answer 29

um idk if its gunna help :/

Answer 30

Nope, so in the denominator: (1 + 1/e^x )^2
Let u = e^x

Answer 31

oooooooooooohhhh lemme try that.. thankyou sooo muchhhhh..

Answer 32

My pleasure :D

Answer 33

i end up with integral of 1/(1+1/u)^2 du?

Answer 34

= Int ( u^2 du / ( u +1 ) ^2

Answer 35

huh? o.O

Answer 36

1 + 1/u = ( u +1 )/ u

Answer 37

umm hold ... you substituted the top with "u" but where did "du" go? du is e^x dx

Answer 38

From (1 + 1/ e^x )^2, take u = e^x
=> du = e^xdx ( on the top)

Answer 39

ohh okay so its integral u^2/(u+1)^2 du

Answer 40

Yup!

Answer 41

now we make another substitution?

Answer 42

no non o

Answer 43

nvm

Answer 44

nvm???

Answer 45

nevermind lol

Answer 46

I thought some technical term :P

Answer 47

lol haha, hey i made another substitution, is that right?

Answer 48

Seem like it !

Answer 49

i took v=1-u: dv=du and u=v-1

Answer 50

and then i opened the square on the top

Answer 51

Should we use partial ?

Answer 52

(v-1)^2/v^2= integral v^2-2v+1/v^2; v^2/v^2-2v^v+1/v^2 dv?

Answer 53

last part v^2/v^2-2v/v^2+1/v^2 dv

Answer 54

Let see if my answer is same as yours!

Answer 55

ok

Answer 56

e^x+3-1/2(e^x+1)^2+c idk :/

Answer 57

I have 3 parts, but stuck at the second:
Int ( udu / (u + 1)^2 )

Answer 58

omg i hate calculus, its fine chloriphyllthankyou very much i got some idea. you helped me ALOT.. thankyou again.. GOD bless you.. i got some more problems. to do also.. :/

Answer 59

Okie, I'll get back to it later!
Just move on to the next one :(

Answer 60

Now I hate e^-x awfully !

Answer 61

lol, m sorry :/

Answer 62

Just move to the next one! I'm pretty sure some higher level OS person will come to rescue!

Answer 63

The last choice is input into wolfram, though!

Answer 64

btw do you how to do those surface area problems? the one with this formula= integral 2pi f(x) sqrt(1+(f ' (x))^2 ?

Answer 65

Just open the new post!

Answer 66

okie dokie

Answer 67

@bmp Can you take a look at this post, see if you can help?

Answer 68

Do you need help with what integral? Or with the DE itself?

Answer 69

The integral

Answer 70

udu / (u + 1)^2 This?

Answer 71

pleaseee

Answer 72

dy = e^xdx / (1+e^-x)^2

Answer 73

I'm stuck with this integral !

Answer 74

Make u = e^x, du = e^x dx, then the integral becomes: \[ \int \frac{du}{(1/u + 1)^2} \]From there, it's a long division, and integrating.

Answer 75

Should we convert into Int ( u^2 du / (u +1)^2 )

Answer 76

Sorry, my browser crashed twice. Not exactly that, notice that we get e^(-x) on the bottom, and we have e^x on top. Not the same thing.

Answer 77

It should ended up with what I posted above, because if u = e^x, then 1/u = e^-x

Answer 78

I did get to that result, but I don't know how to proceed further!

Answer 79

Do a long division of that, you should get something like 1 - 2/(u + 1) + 1/(u+1)^2. :-)

Answer 80

Expand the denominator out and do partial fractions. I think that should do it too.

Answer 81

I seem go around with the result :(

Answer 82

Maybe I should have made myself clearer, 1/(1/x + 1)^2 = x^2/(x^2 + 1)^2 I think it's easier to see it now (the division, I mean).

Answer 83

Yep, that's what I came up!
Int ( u^2 du / ( u +1 ) ^2 )

Answer 84

Do the polynomial long division now. What is x^2/(x^2 + 1)^2? :-)

Answer 85

Or partial fractions, both work just fine :-)

Answer 86

I thought the inside is ( x +1 )^2

Answer 87

Ah, my bad. You are correct, it's only (x+1)^2.

Answer 88

A ( x +1) + ( Bx + C ) = x^2

Answer 89

Yup, albeit I am a bit rusty with my partial fractions :-(

Answer 90

I think x = -1 seems a good candidate.

Answer 91

Does it look right?
A + B = 0
A + B + C = 0

Answer 92

Hmmm, yeah, I got a similar problem here.

Answer 93

Oh, I set up the partial wrong!

Answer 94

x^2 / (x + 1)^2 = A/ ( x+1) + B ( x + 1) ^2

Answer 95

-> A ( x+1) + B = x^2

Answer 96

Shouldnt a +C show up also? The long division yielded a +1 elsewhere.

Answer 97

A C/(x+1)^2, that is.

Answer 98

Cheating a bit: http://www.wolframalpha.com/input/?i=%28x%5E2%29%2F%28x%2B1%29%5E2+partial+fractions