What is the general solution of the differential equation  (y^2+1)dy/dx+2xy^2=2x ?

Sep 19, 2017

$\ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$

Explanation:

We have:

$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{2} = 2 x$ ..... [A]

We can rearrange this non-linear First Order differential equation [A] as follows:

$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 2 x {y}^{2}$
$\therefore \left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \left({y}^{2} - 1\right)$
$\therefore \frac{{y}^{2} + 1}{{y}^{2} - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$

This is now separable, so we can "seperate the variables" to get:

$\int \setminus \frac{{y}^{2} + 1}{{y}^{2} - 1} \setminus \mathrm{dy} = \int \setminus - 2 x \setminus \mathrm{dx}$ ..... [B]

The RHS integral is standard, and the LHS will require a little manipulation, as follows:

$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus \frac{{t}^{2} - 1 + 2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{\left(t + 1\right) \left(t - 1\right)} \setminus \mathrm{dt}$

We can now decompose the fractional part of the integrand into partial fractions, as follows:

$\frac{2}{\left(t + 1\right) \left(t - 1\right)} \equiv \frac{A}{t + 1} + \frac{B}{t - 1}$
$\text{ } = \frac{A \left(t - 1\right) + B \left(t + 1\right)}{\left(t + 1\right) \left(t - 1\right)}$

$2 \equiv A \left(t - 1\right) + B \left(t + 1\right)$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $t = - 1 \implies 2 = - 2 A \implies A = - 1$
Put $t = + 1 \implies 2 = + 2 B \implies B = + 1$

So using partial fraction decomposition we have:

$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus 1 - \frac{1}{t + 1} + \frac{1}{t - 1} \setminus \mathrm{dt}$

Using this result we can now integrate [B] as follows:

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

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

$\therefore - y + \ln | y + 1 | - \ln | y - 1 | = {x}^{2} + C$

$\therefore \ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$

Which, is the General Solution .

We are unable to find a particular solution, as requested, as noi initial conditions have been provided to allow the constant $C$ to be evaluated.