Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

this is a second order y-absent DE

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

*dy/dx

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh wait sorry

OpenStudy (unklerhaukus):

yeah i typed it incorrectly

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

yes, i made a number of errors

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!