Question

Solve the DE\[yy''+(y')^2=0\]

Answer 1

the equation is x-absent

Answer 2

\[\text{let }\frac{\text d y}{\text dx}=p\]\[\frac{d^2y}{dx^2}=p\frac{dp}{\text d y}\]

Answer 3

\[yp\frac{\text d p}{\text dy}-p^2=0\]

Answer 4

\[\frac{\text dp}{\text dy}-\frac1yp=0\]

Answer 5

\[I.F. (y)=e^{\int \frac 1y \text dy}=e^{\ln y}=y\]

Answer 6

\[\frac{\text d}{\text dy}\left(py\right)=0\]\[p=0\]

Answer 7

???

Answer 8

is the ans y=ae^(bx)

Answer 9

the back of my book neglect this particular question for some reason,
how did you arrive at\[y=ae^{bx}\]?

Answer 10

dividing both sides by y^2 we have y'=by and then move on further ....
by the way if u don't mind name/author of the book?

Answer 11

that is the chapter of the text book i am working on and the solutions chapter

Answer 12

and this question i am working on at the momnet is from te problem set 3A 1.g)

Answer 13

thanx...

Answer 14

ur questions are different -------yy′′−(y′)2=0
may be just check the sign whether its - or +

Answer 15

Its just a booklet the Uni produces { University of Wollongong }
the subject is MATH202 Differential Equations II
i could up load any chapter
or the entire text if you want

Answer 16

your right

Answer 17

yy′′+(y′)2=0
might be u have posted a different question ...

Answer 18

i just changed it

Answer 19

\[yy''+(y')^2=0\]

Answer 20

sorry about that

Answer 21

the way u were doing was correct ...
it reduces to yy'=c
and then i hope u can further solve...

Answer 22

sometimes error happens
no need to worry

Answer 23

\[y\times 0=0?\]

Answer 24

yeah the second mistake i made way to leave out the negative sign in the integrating factor step

Answer 25

integrating constant is missing...

Answer 26

\[-p=c\]
\[?

Answer 27

didn't get ...

Answer 28

I get so confused
in that chapter
under section x-absent page 1
it says ..."...then,
\[p=\frac{\text dy}{\text dx}=\frac{\text d^2y}{\text d x^2}\]"
is this right?

Answer 29

\[ (yy')' = 0 \]
\[ yy' = k_1\]
\[ \frac{y^2}{2} = k_1x + k_2\]

Answer 30

i don't agree with the textbooks statement
\[p=\frac{dy}{dx}=\frac{d^2y}{dx^2}\]
since:
\[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=\frac{dp}{dy}y^'=p \frac{dp}{dy}\]

Also, the equation is seperable where you do IF.

Answer 31

i thought it was a typo too but i wasent sure

Answer 32

looks like a pretty decent online text though

Answer 33

it is from the booklet i have it in hard copy too

Answer 34

two versions actually i failed this subject

Answer 35

sorry i left in between that as i had some work....
i hope ur answer is solved