Solve the differential equation xyy' +xyy'= y^2 +1?

Mar 18, 2017

$y = \pm \sqrt{B x - 1}$

Explanation:

We have:

$x y y ' + x y y ' = {y}^{2} + 1$
$\therefore 2 x y y ' = {y}^{2} + 1$
$\therefore \frac{y}{{y}^{2} + 1} y ' = \frac{1}{2 x}$

This is a First Order separable DE, so we can separate the variables to get;

$\int \setminus \frac{y}{{y}^{2} + 1} \setminus \mathrm{dy} = \int \setminus \frac{1}{2 x} \setminus \mathrm{dx}$

The RHS is trivial and for the LHS we can use a substitution:
Let $u = {y}^{2} + 1 \implies \frac{\mathrm{du}}{\mathrm{dy}} = 2 y$, or $\int \frac{1}{2} \ldots \setminus \mathrm{du} = \int \ldots y \setminus \mathrm{dy}$

Substituting we get

$\int \setminus \frac{\frac{1}{2}}{u} \setminus \mathrm{du} = \int \setminus \frac{1}{2 x} \setminus \mathrm{dx}$

We can now integrate to get:

$\frac{1}{2} \ln | u | = \frac{1}{2} \ln | x | + A$
$\therefore \ln | u | = \ln | x | + 2 A$
$\therefore \ln | u | = \ln | x | + \ln B$ (say)
$\therefore \ln | u | = \ln B | x |$
$\therefore u = B x$
$\therefore {y}^{2} + 1 = B x$
$\therefore {y}^{2} = B x - 1$
$\therefore y = \pm \sqrt{B x - 1}$