Is My Technique Correct for Solving 2nd Order Nonlinear Differential Equations?

[j_reez]

this is the problem: xy'' -x(y')^2 = y'

my book says that i need to substitute u=y' and du/dx=y''...
so i get:

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

so next the book says i need to separate the x's/dx' to one side and u's/du's to the other. however, i cannot do it

am i using the correct technique?


thanks-
Justin

 

[Galileo]

edit: Nevermind

 

[j_reez]

this problems driving me crazy

 

[ehild]

this is the problem: xy'' -x(y')^2 = y'

my book says that i need to substitute u=y'

Substitute xy'=u instead.

ehild

 

[j_reez]

i don't see how that would work...if u=xy' then du/dx=y' +xy''
how would thos substitutions work?

 

[wisky40]

if you want you can work it like this:

x(du/dx)-xu^2=u => (u^-2)(du/dx)-(u^-1)/x=1 then use this

substitution, I think this is "Bernoulli equation" z=-(u^-1) and dz/dx=(u^-2)(du/dx)

=> dz/dx +z/x=1. I hope this help you.

 

[James R]

Your equation is:

xy'' -x(y')^2 = y'

Since y does not appear in the equation, start by putting:

y' = u

which means the equation is:

xu' - xu^2 = u
u' - u^2 = \frac{u}{x}

Now put

u(x) = xv(x)

So that

u' = v'x + v

The differential equation then becomes separable, as follows:

v'x + v - v^2 x^2 = v
xv' = v^2 x^2
dv/v^2 = x dx

Integrating gives:

-\frac{1}{v} = \frac{x^2}{2} + c

where c is an undetermined constant. Rewriting v in terms of u gives:

-\frac{x}{u} = \frac{x^2}{2} + c

Rearranging, we get:

-\frac{u}{x} = \frac{1}{x^2/2 + c}

or

y' = -\frac{x}{x^2/2 + c}

This can now be easily integrated to give the solution:

y = -\ln(x^2/2 + c) + d

where d is a second constant of integration.

 

[ehild]

i don't see how that would work...if u=xy' then du/dx=y' +xy''
how would thos substitutions work?

y' = \frac{u}{x}\mbox { , } y''=\frac{u'x-u}{x^2}

Plugging in y' and y'' the differential equation becomes

\frac{xu'-u}{x}-\frac{u^2}{x}=\frac{u}{x}\rightarrow u'x=u^2+2u

This is easily separable:

\frac{du}{u^2+2u}=\frac{dx}{x}

and results in

u=\frac{2x^2}{C-x^2} \mbox { that is } y'=\frac{2x}{C-x^2}

This is the same solution James got but his method is simpler, I have to admit You see there are many ways to go...

ehild

 

[j_reez]

Thank You

thanks guys! its very clear to me now :) my book had one very limited example which made it sound like i only need to make one substitution. however, i clearly see it now!

thanks,
Justin