Question

\[y''-yy'=0\]

Answer 1

the equation is x-absent
let \[y'=p\]\[y''=pp'\]

Answer 2

\[pp'-yp=0\]\[p'-y=0\]\[dp/dx=y\]

Answer 3

?

Answer 4

whats your question?

Answer 5

what do i do next
i am tring to solve for y

Answer 6

y=y(x)

Answer 7

is p a constant?

Answer 8

p=dy/dx

Answer 9

y'=p
=>y''=0

Answer 10

if p is constant

Answer 11

y''=pp'

Answer 12

p must be a const just check it out!

Answer 13

that is just the trivial solution

Answer 14

k

Answer 15

the other solution has tangent of something in it

Answer 16

y'=p
y''=p'

Answer 17

ur solving for y right

Answer 18

?

Answer 19

i am solving for y(x) yes.
y'=p
however
y''=pp'

Answer 20

then you should integrate this equation y'=p
=> y(x)=px+c

Answer 21

y(x)=y'x+c

Answer 22

err p is a function of x dhashni

Answer 23

I am tring to show that the solution is \[y=c\] or \[y=atan({ax \over 2}+b)\]

Answer 24

rather \[y = a \tan(\frac{ax}{2}+b)\]

Answer 25

um...

Answer 26

i dont know

Answer 27

aha - it goes tan -> sec^2 -> tan*sec^2 as you differentiate

Answer 28

wait why do you think that y'' = pp' rather y'' = yy'

Answer 29

\[y'' - yy' = 0\]\[y' - \frac{1}{2}y^2 = c\]\[y' = \frac{1}{2}y^2 + c\]\[\frac{y'}{\frac{1}{2}y^2 + c} = 1\]\[\frac{2y'}{y^2 + 2c} = 1\]\[\sqrt{\frac{2}{c}}\arctan(\frac{y}{\sqrt{2c}})=x+c_2\]\[y = \sqrt{2c} \tan(\sqrt{\frac{c}{2}}(x+c_2))\]

Answer 30

the two hard steps are where you integrate both sides with respect to x

Answer 31

hey thankyou very much

Answer 32

nps :)

Answer 33

i can not see what you did to
\[y''-yy'=0\]to get
\[y'-{1 \over 2}y^2=c\]

Answer 34

that's the step where i integrated. the integral of y'' is y'+c and the integral of 2yy' is y^2+c (chain rule)

Answer 35

\[\int (y''-yy') dx = \int 0 \space dx\]
\[\int \frac{d}{dx}(y') dx - \frac{1}{2} \int 2y\frac{dy}{dx} dx = 0 + C_3\]
\[ y' + C_1 - \frac{1}{2}y^2 - C_2 = C_3 \]
\[ y' - \frac{1}{2}y^2 = (-C_1+C_2+C_3) \]
\[ y' - \frac{1}{2}y^2 = c \]

Answer 36

that is it
thank you Broken Fixer