Question

What is the general solution of the differential equation `xy'' = y'+x(y')^2`?

Answer 1

`y = C-ln |x^2+B|`

Explanation:

We have:

`xy'' = y'+x(y')^2` ..... [A]

Which is a Second Order Non-Linear ODE. As there is no term in `y` then let us attempt the following substitution:

`v = y' => v'=y''`

Substituting into the DE [A] we get:

`xv' = v+xv^2` ..... [B]

Which has reduced the initial ODE to a First Order Non-Linear ODE (in this case a Bernoulli Equation), so now we attempt a second substitution of the form:
`u = v^(-1) => (du)/(dv) = -v^(-2)` and `(dv)/(du) = -v^2`

Substituting into the last DE [B], in conjunction with the chain rule we get;

`x(dv)/(dx) = v+xv^2`
`:. x(dv)/(du) * (du)/(dx) = v+xv^2`
`:. x(-v^2) * (du)/(dx) = v+xv^2`
`:. - (du)/(dx) = 1/(xv)+1`
`:. - (du)/(dx) = u/x+1`
`:. (du)/(dx) + u/x = -1` ..... [C]

So the second substitution has reduced the DE into a first order linear differential equation of the form:

`(d zeta)/dx + P(x) zeta = Q(x)`

We solve this using an Integrating Factor

`I = exp( \ int \ P(x) \ dx )`
`\ \ = exp( int \ (1/x) \ dx )`
`\ \ = exp( lnx )`
`\ \ = x`

And if we multiply the last equation [C] by this Integrating Factor, `I`, we will have a perfect product differential;

`\ \ \ x(du)/(dx) + u = -x`
`:. -d/dx ( xu ) = x`

Which is now a trivial separable DE, so we can integrate to get:

`-xu = int \ x \ dx`
`\ \ \ \ \ \ \ = x^2/2 + K`

And restoring the `u` substitution we have:

`-x(1/v) = x^2/2 + K`
`:. -x/v = (x^2+2K)/2`
`:. -v/x = 2/(x^2+A)`
`:. v = -(2x)/(x^2+A)`

And restoring the `v` substitution we have::

`y' = -(2x)/(x^2+B)`

Which again is a First Order separable ODE, so again we "separate the variable" to get:

`int \ dy = -int \ (2x)/(x^2+B) \ dx`

Integrating we get:

`y = -ln |x^2+B| + C`

Which is the GS of [A]