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UnkleRhaukus
 4 years ago
\[y''+x(y')^2=0\]
UnkleRhaukus
 4 years ago
\[y''+x(y')^2=0\]

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UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0this is a second order yabsent DE

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0let\[p=y'\]\[p'=y''\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[p'+xp^2=0\] \[p'=xp^2\] \[dp/p^2=xdx\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int{dp \over p^2}=\int xdx\]\[1/p+c_1=x^2/2\] \[1/p=x^2/2c_1\] \[dy/dy={ 1\over 2}{ 1 \over x^22c_1 }\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[y={1 \over 2} \int{1 \over x^2c_2^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i typed it incorrectly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would be y' = 2/x^2  1/c1

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yes, i made a number of errors
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