- #1
j_reez
- 9
- 0
problem: xy'' -x(y')^2 = y'
what i have so far:
u=y' and du/dx=y''
du/dx - u^2 = (1/x)u
int[(1/u)-u]du = int[1/x]dx
ln u - (1/2)u^2 = ln x +c
ok, now is what I've done so far correct? what do i do next?
ps: i'd like to say hi to everyon :) I am new here
what i have so far:
u=y' and du/dx=y''
du/dx - u^2 = (1/x)u
int[(1/u)-u]du = int[1/x]dx
ln u - (1/2)u^2 = ln x +c
ok, now is what I've done so far correct? what do i do next?
ps: i'd like to say hi to everyon :) I am new here