Question

diff equations

Answer 1

\[yy"-(y')^2+y'=0\]
\[\int y\frac{dy'}{dx}dx-\int y'\frac{dy}{dx}dx+\int\frac{dy}{dx}dx=0\]
\[yy'-\int y'\frac{dy}{dx}dx-yy'+\int y\frac{dy'}{dx}+y=0\]

Answer 2

I'm sure that you're capable of continuing from here

Answer 3

I know that there are 3 cases when dealing second order, right? =)

Answer 4

I don't really think I'm following the usual way :)

Answer 5

You have your own technique in analyzing second order?

Answer 6

Correction:
\[\int y\frac{dy'}{dx}dx-\int y'\frac{dy}{dx}dx+\int\frac{dy}{dx}dx=c_1\]

Answer 7

Wow. =)

Answer 8

well this ain't the usual second orders

Answer 9

You mean..

Answer 10

Im lost..

Answer 11

Well, ain't the usual second orders in the form of ay"+by'+cy=0?

Answer 12

Right..

Answer 13

I just integrated the whole thing

Answer 14

How do i integrate y' dy/dx dx? Omg this is so so basic.. Im sorry

Answer 15

well, integration by part (\(\displaystyle\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx}dx\))

Answer 16

u = y' then du = dy/dx

Answer 17

well, u=y' then du=dy'/dx

Answer 18

dv= dv/dx then v= v' ?

Answer 19

\[yy"-(y')^2+y'=0\]
First part:
\[\int yy'' dx =\int y\frac{dy'}{dx}dx=yy'-\int y'\frac{dy}{dx}dx\]
Second part:
\[\int y'y' dx =\int y'\frac{dy}{dx}dx=yy'-\int y\frac{dy'}{dx}dx\]
Third part:
\[\int y' dx =\int \frac{dy}{dx}dx=\int dy\]

Answer 20

understand?

Answer 21

Well honestly.. I understand the first part.. Only..

Answer 22

Oops, I guess my approach doesn't work

Answer 23

well if you can get the first part i believe that you can also get the second part as well

Answer 24

I think so.. Yeah. I can't reach you. Haha =)

Answer 25

No, i dont get your idea on the second part

Answer 26

basically just writing y' as dy/dx

Answer 27

Yes yes.

Answer 28

Oh no.