Question

solve this nonlinear differential equation:
y''+(y'^2)+1=0
please show your work

Answer 1

so y' is squared?

Answer 2

yes it is

Answer 3

so its (y')^2

Answer 4

yes thats right

Answer 5

ok let me think

Answer 6

al right no problem but it would be helpful if you have had a course in differential equations otherwise it will be really challenging

Answer 7

here is a hint...this involves using u=y' to reduce it to first order equation but it's tricky

Answer 8

i think i got it

Answer 9

one sec let me look at it to make sure

Answer 10

ok no problem take your time you will get medal for this...promise

Answer 11

i kind of want to type what i did but its wrong maybe someone can say where i went wrong

Answer 12

i know how to work the problem...i have it done but don't know if its right because my answer does not match with the solution of this problem...we can help each other help don't worry just post what you have..

Answer 13

dy/dx=v
dv/dy=F(y,v)
d^2y/dx^2=dv/dx=dv/dy(dy/dx)=vdv/dy

Answer 14

i just need to know what you got for dy/dx ?

Answer 15

you need to do
Vdv/dy=-V^2-1

Answer 16

i get dy/dx=(e^(2c1-2y)-1)^(1/2)

Answer 17

V/V^2-1dv=dy

Answer 18

noxi i am sorry i kind of do not understand what you are saying...are you differentiating?

Answer 19

1/2ln(V^2-1)=y+C

Answer 20

ok sorry i remember doing this last year its a special case when you only have one variable

Answer 21

i get 1/2ln(v^2+1)=-y+c

Answer 22

its called second order equations with independent variable missing

Answer 23

my u is ur v...

Answer 24

ur on the right track noxi but seems like you missed a sign somewhere but keep going and post the solution y now

Answer 25

also let me know what you get for your dy/dx before you solve for y

Answer 26

lets see
1/2ln(V^2-1)=-y+C
ln(v^2-1)=-2(y)+C
(v^2-1)=e^(-2y+C)
v^2=e^(-2y+C)-1
dy/dx=SQRT(e^(-2y+C)-1)

Answer 27

so yeah i got the same thing

Answer 28

seems good so far...just solve for y

Answer 29

aya are you there?

Answer 30

yes dy/dx=SQRT(e^(-2y+C)-1)
\[dy/\sqrt(e^(-2y+C)-1)=dx\]
\[dy/\sqrt(C1e^(-2y0-1)=dx\]
sin^-1(e^-y)=x?

Answer 31

no no no no never mind forget that last part that was stupid

Answer 32

shoshny and mandolino what are you guys talking about?

Answer 33

aya can you post whatever you have so far?

Answer 34

MeEngr what are you gettinG?

Answer 35

shoshny i don't know what you talking about man

Answer 36

this looks like an ugly integral
\[\int\limits_{}^{}dy/\sqrt(e^(-2y)-1)\]

Answer 37

yes it does very ugly indeed...i cant even type here here but it has natural log function in it

Answer 38

=x i would leave it like that lol

Answer 39

i typed it in wolfram and theres imaginarys in there

Answer 40

yea true but you noxi whats weird is that if you solve this nonlinear differential equation on wolfram http://www.wolframalpha.com/input/?i=y%27%27%2B%28y%27^2%29%2B1%3D0 you get a really simple answer but it does not match with what we got so i am confused...is wolfram right?

Answer 41

Vdv/dy=-V^2-1 do we agree up to heRE/

Answer 42

i agree with u all the way up to dy/dx we are right i believe so whatever we do after that is also right but why is the solution from wolfram so different its really weird..

Answer 43

did you look at the solution in wolfram?

Answer 44

aya can u please post what you have so far?

Answer 45

dv/dy=-(v+(1/v))
dv/(v+1/v)=-dy
dy=-dv/(v^2+1/V)
dy=-vdv/(v^2+1)
y=-1/2ln(V^2+1)
e^2y=1/v^2+1
1/e^2y-1=v^2
(1-e^2y)/e^2y=v^2
sqrt(1-e^2y)/e^2y=v=dy/dx

Answer 46

noxi...what did you substitute for y' and y''?

Answer 47

i think
dx=sqrt(e^2y)dy/sqrt(e^2y-1)
now integrate and wolfram alpha gave a do able integral check me on my math to make sure thats correct is http://www.wolframalpha.com/input/?i=sqrt%28e%5E2y%29%2Fsqrt%28e%5E2y-1%29

Answer 48

i think something got screwed up in your calculations...

Answer 49

isnt the prob is like this?
yy''+(y'^2)+1=0 here the answer is
equation of circle
(x- 1/2)^2+(y+ 1/4)^2 =1/16 answer

Answer 50

noxi answer my previous question

Answer 51

i subbed y'=dy/dx=V
y''=dv/dx=dv/dy(dy/dx)=Vdv/dy

Answer 52

thats correct noxi...but your dy/dx looks fishy

Answer 53

which one? when i did my calculations

Answer 54

i think we did this correctly and the integral is horrible, i am off to bed, good luck and sorry i couldnt get the integral right

Answer 55

y+c=-1/2ln(V^2+1)
-2y-2c=ln(V^2+1)
e^(-2y-2c)=sqrt(V^2+1)

Answer 56

you had a problem with constant of integration and sign...but thanks for help appreciate it

Answer 57

Ce^(-2y)=e^(-2y-2c)

Answer 58

your sqrt(is in the wrong PLace)

Answer 59

yea but here we can not bring the constant in front of e because of other problems involving boundary conditions it does not stay consistent but other than that and a few sign problems your right

Answer 60

i wanted to confirm i am headed the right direction and you made me feel that i am working this right...so thanks for showing me that..

Answer 61

hey aya can you send me a message in group chat...i am just wondering if you have any solution yet

Answer 62

mark i am sorry about late reply but how did you arrive at that solution?

Answer 63

no i'm still working on it
i found my mistake with that one way

Answer 64

ok can i see how far you have gotten...

Answer 65

i will only show something if i have something good to show

Answer 66

and i do not yet

Answer 67

your welcome sorry i am now off to bed, good luck and problems like these on tests arent this ugly.

Answer 68

isnt the prob is like
yy'+(y')^2+1=0
yy'+(y')^2=-1 MULT BY 2
2[yy'+(y')^2]=-2
2+2[yy'+(y')^2]=0
2+0+2[yy'+(y')^2]=0 integrating
2x-1+0+2yy'=0 integrating
x^2-x+ 1/4 +y^2= 1/16
(x- 1/2)^2+(y+ 1/4)^2 =1/16 answer

Answer 69

sory m pc kept hanging up on me..lol

Answer 70

ME hope thats what you are looking for....good luck now

Answer 71

mark what happened to y''?

Answer 72

This is pretty straight forward by substituting u = y'

Then
\[u' + u^2 + 1 = 0 \]
which is a separable equation in u. Solve for u then integrate once to obtain y.

Or am I missing something?

Answer 73

no you need to sub y'' as well

Answer 74

if u=dy/dx
du/dx=du/dy(dy/dx)=udu/dy

Answer 75

If u = y', then u' = y'' and the equation is
\[u' + u^2 + 1 = 0\]
Separating variables
\[du/(u^2+1) = -dx\]
and integrating
\[\tan^{-1} u = - x + c_1 \ \hbox{ namely } \ u(x) = \tan(-x+c_1)\]
Now,
\[dy/dx = u \ \hbox{ hence } y = \int\limits \tan(-x+c_1) \ dx = \ln (\cos (-x+c_1)) + c_2\]

It is easy to verify this is a solution to the equation:
\[y' = \tan (-x+c_1) \ \hbox{ and} \ y'' = -\sec ^2(-x+c_1)\]
hence
\[y'' + (y')^2 + 1 = -\sec ^2(-x + c_1) + \tan^2(-x + c_1) + 1 = 0\]
as\[\tan^2 \theta + 1 = \sec ^2 \theta \ , \ \ \hbox{for all } \ \theta\]

Answer 76

oh wait a minute...
you divided both sides by u^2+1

and
so we get
u'/(u^2+1)+1=0
omg brillant

Answer 77

teach me your ways james

Answer 78

Wow, that was amazing!

Answer 79

i looked at this last night and i was trying the only way i know
and was failing and then james comes in shows a freaking delicious way
and i'm envious and jealous at the same time

Answer 80

I was trying by substituting \(y=\frac{u'}{u}\). Not sure of that was gonna work, but I got \(u''+u=0\) which is very easy to solve and then substitute back.

Answer 81

Ha, well. I'm glad I've still got it. *cracks knuckles* Next?

Answer 82

ok i'm going to appoint james as my second zarkon

Answer 83

I fanned you James! Not many people get that :P

Answer 84

lol i fanned james too

Answer 85

good job james, turning a nonlinear eq into a linear eq.

great eye

Answer 86

Not quite:\[ u' + u^2 + 1 = 0 \]is not linear, but it is separable.

Answer 87

thats what i meant sep not linear

lets see james no one else answered this can you find the annihilator for sin(x)e^(-x)
i know itS D^2-2D+A^2+B^2 but i wasnt getting it to =0