# What is the particular solution of the differential equation? :  dx/(x^2+x) + dy/(y^2+y) = 0  with y(2)=1

Jul 5, 2017

$\left(1\right) : \frac{x y}{\left(x + 1\right) \left(y + 1\right)} = c , \text{ is the GS.}$

$\left(2\right) : 2 x y = x + y + 1 , \text{ is the PS.}$

#### Explanation:

The given Diff. Eqn. $\frac{\mathrm{dx}}{{x}^{2} + x} + \frac{\mathrm{dy}}{{y}^{2} + y} = 0 ,$ with

Initial Condition (IC) $y \left(2\right) = 1.$

It is a Separable Variable Type Diff. Eqn., &, to obtain its

General Solution (GS), we integrate term-wise.

$\therefore \int \frac{\mathrm{dx}}{x \left(x + 1\right)} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \frac{\left(x + 1\right) - x}{x \left(x + 1\right)} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \left\{\frac{x + 1}{x \left(x + 1\right)} - \frac{x}{x \left(x + 1\right)}\right\} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \int \left\{\frac{1}{x} - \frac{1}{x + 1}\right\} \mathrm{dx} + \int \frac{\mathrm{dy}}{y \left(y + 1\right)} = \ln c .$

$\therefore \left\{\ln x - \ln \left(x + 1\right)\right\} + \left\{\ln y - \ln \left(y + 1\right)\right\} = \ln c .$

$\therefore \ln \left\{\frac{x}{x + 1}\right\} + \ln \left\{\frac{y}{y + 1}\right\} = \ln c .$

$\therefore \ln \left\{\frac{x y}{\left(x + 1\right) \left(y + 1\right)}\right\} = \ln c .$

$\therefore \frac{x y}{\left(x + 1\right) \left(y + 1\right)} = c , \text{ is the GS.}$

To find its Particular Solution (PS), we use the given IC, that,

$y \left(2\right) = 1 , i . e . , w h e n , x = 1 , y = 2.$

Subst.ing in the GS, we get, $\frac{\left(1\right) \left(2\right)}{\left(1 + 1\right) \left(2 + 1\right)} = c = \frac{1}{3.}$

This gives us the complete soln. of the eqn. :

$\frac{x y}{\left(x + 1\right) \left(y + 1\right)} = \frac{1}{3} , \mathmr{and} , 2 x y = x + y + 1.$

Jul 5, 2017

$y = \frac{x + 1}{2 x - 1}$

#### Explanation:

We have an differential equation equation in the form of differentials:

$\frac{\mathrm{dx}}{{x}^{2} + x} + \frac{\mathrm{dy}}{{y}^{2} + y} = 0$

We can write this in "separated variable" form as follows and integrate both sides

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

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

Now et us find the partial fraction decomposition of $\frac{1}{{u}^{2} + u}$ which we can use on both integrals:

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

$1 \equiv A \left(u + 1\right) + B u$

We can find the constant coefficients $A$ and 'B' vis substitution (effectively the "cover-up method")

Put $u = 0 \setminus \setminus \setminus \setminus \setminus \implies 1 = A$
Put $u = - 1 \implies 1 = - B$

Thus:

$\frac{1}{{u}^{2} + u} \equiv \frac{1}{u} - \frac{1}{u + 1}$

Using this in the above we get:

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

We can now evaluate the integrals (not forgetting the constant of integration) to get:

$\ln | y | - \ln | y + 1 | = - \left\{\ln | x | - \ln | x + 1 |\right\} + c$

Then rearranging and using the properties of logarithms we have:

$\ln | y | - \ln | y + 1 | + \ln | x | - \ln | x + 1 | = c$
$\therefore \ln \left(\frac{| x | | y |}{| x + 1 | | y + 1 |}\right) = c$
$\therefore \ln \left(| \frac{x y}{\left(x + 1\right) \left(y + 1\right)} |\right) = c$
$\therefore | \frac{x y}{\left(x + 1\right) \left(y + 1\right)} | = {e}^{c}$

Now ${e}^{c} > 0 \forall c \in \mathbb{R}$, thus we can remove the modulus operator, giving the General Solution:

$\frac{x y}{\left(x + 1\right) \left(y + 1\right)} = A$, say, where $A > 0$.

We are also given that $y \left(2\right) = 1$, this tells us that:

$\frac{2 \cdot 1}{\left(2 + 1\right) \left(1 + 1\right)} = A \implies A = \frac{2}{3}$

Thus the required solution is:

$\frac{x y}{\left(x + 1\right) \left(y + 1\right)} = \frac{1}{3}$

$\therefore 3 x y = \left(x + 1\right) \left(y + 1\right)$
$\therefore 3 x y = \left(x y + x + y + 1\right)$
$\therefore 3 x y = x y + x + y + 1$
$\therefore 2 x y - y = x + 1$
$\therefore y \left(2 x - 1\right) = 1 x + 1$
$\therefore y = \frac{x + 1}{2 x - 1}$