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

Oct 25, 2017

$y = C - \ln | {x}^{2} + B |$

#### Explanation:

We have:

$x y ' ' = y ' + x {\left(y '\right)}^{2}$ ..... [A]

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

$v = y ' \implies v ' = y ' '$

Substituting into the DE [A] we get:

$x v ' = v + x {v}^{2}$ ..... [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:
$u = {v}^{- 1} \implies \frac{\mathrm{du}}{\mathrm{dv}} = - {v}^{- 2}$ and $\frac{\mathrm{dv}}{\mathrm{du}} = - {v}^{2}$

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

$x \frac{\mathrm{dv}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore x \frac{\mathrm{dv}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore x \left(- {v}^{2}\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore - \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x v} + 1$
$\therefore - \frac{\mathrm{du}}{\mathrm{dx}} = \frac{u}{x} + 1$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} + \frac{u}{x} = - 1$ ..... [C]

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

$\frac{d \zeta}{\mathrm{dx}} + P \left(x\right) \zeta = Q \left(x\right)$

We solve this using an Integrating Factor

$I = \exp \left(\setminus \int \setminus P \left(x\right) \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\int \setminus \left(\frac{1}{x}\right) \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\ln x\right)$
$\setminus \setminus = x$

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

$\setminus \setminus \setminus x \frac{\mathrm{du}}{\mathrm{dx}} + u = - x$
$\therefore - \frac{d}{\mathrm{dx}} \left(x u\right) = x$

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

$- x u = \int \setminus x \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} / 2 + K$

And restoring the $u$ substitution we have:

$- x \left(\frac{1}{v}\right) = {x}^{2} / 2 + K$
$\therefore - \frac{x}{v} = \frac{{x}^{2} + 2 K}{2}$
$\therefore - \frac{v}{x} = \frac{2}{{x}^{2} + A}$
$\therefore v = - \frac{2 x}{{x}^{2} + A}$

And restoring the $v$ substitution we have::

$y ' = - \frac{2 x}{{x}^{2} + B}$

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

$\int \setminus \mathrm{dy} = - \int \setminus \frac{2 x}{{x}^{2} + B} \setminus \mathrm{dx}$

Integrating we get:

$y = - \ln | {x}^{2} + B | + C$

Which is the GS of [A]