# Need help with nonlinear 2nd order DE

1. Nov 8, 2004

### j_reez

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

2. Nov 9, 2004

### TenaliRaman

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

3. Nov 9, 2004

### j_reez

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

4. Nov 9, 2004

### j_reez

ok hows this look:

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

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

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

?

5. Nov 9, 2004

### TenaliRaman

du/dx - u^2 = (1/x)u
Then,
Multiplying throughout by x gives,
x(du/dx) - u^2*x = u

-- AI

6. Nov 9, 2004

### j_reez

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?

7. Nov 9, 2004

### TenaliRaman

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

8. Nov 9, 2004

### j_reez

im really not seeing how this can be separated :grumpy:

9. Nov 9, 2004

### TenaliRaman

The thing is that it cannot be separated atleast in this form....

Substitute u = vx ...
where v is some function of x ...

-- AI