Question

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

Answer 1

this is a second order y-absent DE

Answer 2

let\[p=y'\]\[p'=y''\]

Answer 3

\[p'+xp^2=0\]
\[p'=-xp^2\]
\[-dp/p^2=xdx\]

Answer 4

\[\int-{dp \over p^2}=\int xdx\]\[1/p+c_1=x^2/2\]
\[1/p=x^2/2-c_1\]
\[dy/dy={ 1\over 2}{ 1 \over x^2-2c_1 }\]

Answer 5

*dy/dx

Answer 6

\[y={1 \over 2} \int{1 \over x^2-c_2^2}\]

Answer 7

if 1/p is y' wouldnt it either be dx/dy or wouldnt the 1/2 also be inversed. you put the x^2 and c1 on the bottom but didnt change the 1/2

Answer 8

oh wait sorry

Answer 9

yeah i typed it incorrectly

Answer 10

it would be y' = 2/x^2 - 1/c1

Answer 11

yes, i made a number of errors