## need help with nonlinear 2nd order DE

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 ive done so far correct? what do i do next?

ps: i'd like to say hi to everyon :) im new here
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 j_reez, welcome to the board! You have made an error here --> du/dx - u^2 = (1/x)u int[(1/u)-u]du = int[1/x]dx What u will have is ... du/dx - u^2 = (1/x)u du - u^2dx =(u/x)dx Can u see the error u made? Can u correct it? -- AI
 i must be making a trivial algebraic mistake....as far as i know im supposed to be isolating dx's and x's on one side with u's and du's on the other...which is why i divided through by x. is this not allowed? oh boy, i see it...i cant get dx to the other side like that...let me see what i can do

## need help with nonlinear 2nd order DE

ok hows this look:

x(du/dx) -u^2 = u

x(du/dx) = u + u^2

(1/x)dx = (1/u+u^2)du

?
 du/dx - u^2 = (1/x)u Then, Multiplying throughout by x gives, x(du/dx) - u^2*x = u -- AI
 yes that was the form it was in. ive got it down to this: [int]dx/x = [int]du/(u(u+1)) how do i integrate the RHS?
 j_reez, think again ... ur original equation was, xy'' -x(y')^2 = y' placing u = y' u get, xu'--xu^2 = u The way u have separated wont work ... Think harder!! :) -- AI P.S as an aside, to integrate 1/(u(u+1)) u should use partial fractions and integrate ofcourse for now it wont apply to this problem i am just telling this as it might be helpful somewhere else
 im really not seeing how this can be separated
 The thing is that it cannot be separated atleast in this form.... Substitute u = vx ... where v is some function of x ... -- AI