Question

yy''-y^'2+y'=0 general solution?

Answer 1

Do you mean
\[ yy'' - (y')^2 + y' = 0 \] ?

Answer 2

yes

Answer 3

wolframalpha suggests a nice substitution that makes the differential equation a separable one.

Answer 4

Notice first that
y y'' = y' (y' - 1)
hence
\[ y'' \left( \frac{1}{y'} - \frac{1}{y'-1} \right) = \frac{1}{y} \]
Now integrate once, find a new equation involving only y' and y; and then solve again.

Answer 5

I flipped the signs on the LHS.

Answer 6

thanks

Answer 7

got it

Answer 8

y'=P ,y''=PdP/dy,so P(ydP/dy-P+1)=O ..........

Answer 9

Here's an elegant solution. Divide through by y^2 and note that
\[ \left(\frac{y'}{y}\right)' = \frac{yy'' - (y')^2}{y^2} \ \ \hbox{ and } \ \ \left(\frac{1}{y}\right)'= -\frac{y'}{y^2} \]

Answer 10

Very smart!