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\[y''+x(y')^2=0\]
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this is a second order y-absent DE
let\[p=y'\]\[p'=y''\]
\[p'+xp^2=0\] \[p'=-xp^2\] \[-dp/p^2=xdx\]
\[\int-{dp \over p^2}=\int xdx\]\[1/p+c_1=x^2/2\] \[1/p=x^2/2-c_1\] \[dy/dy={ 1\over 2}{ 1 \over x^2-2c_1 }\]
*dy/dx
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\[y={1 \over 2} \int{1 \over x^2-c_2^2}\]
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
oh wait sorry
yeah i typed it incorrectly
it would be y' = 2/x^2 - 1/c1
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yes, i made a number of errors
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