Question

y^2*y''=y'

Answer 1

\[y^2.y''=y'\]

Answer 2

i substituted: y'=P and y''=dP/dy * P
solve for P: P=-1/y + c

Answer 3

then i'm lost

Answer 4

After substituting, you should have
\[y^2 P(y)'P(y)=P(y(x))\\y^2P(y)'P(y)-P(y) =0\\P(y)*[y^2P'-1]=0\]

Answer 5

--> either P(y)=0 or y^2P' -1 =0
1) P(y) =0 --> ...
2) y^2P' -1 =0 solve for y ..
can you continue?

Answer 6

i will have to check it out tomorrow

Answer 7

but thanks

Answer 8

ok

Answer 9

the answer should be \[\frac{ 1 }{ c }y + \frac { 1 }{c^2}\ln \left| -1+cy \right| = x+c\]

Answer 10

right?

Answer 11

btw, what you said, i already did solve it: i got P=-1/y + c

Answer 12

couldn't solve for y, because i need P first, right?

Answer 13

2) y^2P' -1 =0 => P=-1/y + c

Answer 14

\[P=\frac{ dy }{ dx }=\frac{ 1-cy }{ y }\]
separate the variables and integrate
\[\int\limits \frac{ y }{ 1-cy }dy=\int\limits dx +c1\]
\[\frac{ -1 }{ c } \int\limits \frac{ 1-c y-1 }{ 1-cy } dy=x+c2\]
i think now you can complete.

Answer 15

correction
write c1 in place of c2

Answer 16

your last step, could you explain that please?
I see alot of people doing that, i don't quite understand why

Answer 17

\[\frac{ y }{ 1-cy }=\frac{ -1 }{ c }*\frac{ -cy }{ 1-cy }\] (multiply and divide by -c
add 1 and subtract 1 in the numerator
\[=\frac{ -1 }{ c }\times \frac{ 1-cy-1 }{ 1-cy }=\frac{ -1 }{ c}\left( 1-\frac{ 1 }{ 1-cy } \right)\]
or simply divide by long division

Answer 18

when the power of numerator is greater than or equal to the denominator ,then we generally divide.

Answer 19

can you complete it?