# What is the general solution of the differential equation?  y^2 \ dx + xy \ dy = 0

## What can you say about the solutions?

Dec 10, 2017

Solution of this differential euation was $\frac{1}{2} {x}^{2} {y}^{2} = C$

#### Explanation:

$M \left(x , y\right) = {y}^{2}$ and $N \left(x , y\right) = x y$

Once, I controlled condition exactness of differential equation,

$\frac{\mathrm{dM}}{\mathrm{dy}} = 2 y$ and $\frac{\mathrm{dN}}{\mathrm{dx}} = y$

Hence, it wasn't exact.

I decided to integration factor accordingly $x$

Hence,

P=e^[int ((dM)/dy-(dN)/dx)/N*dx

=e^[int (2y-y)/(xy)*dx

=${e}^{\int \frac{\mathrm{dx}}{x}}$

=${e}^{L n x}$

=$x$

After multiplying both sides with $P = x$,

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

After controlling exactness condition of it,

$\frac{d \left(x {y}^{2}\right)}{\mathrm{dy}} = \frac{d \left({x}^{2} y\right)}{\mathrm{dx}} = 2 x y$

Thus, solution of it,

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

$d \left(\frac{1}{2} {x}^{2} {y}^{2}\right) = 0$

$\frac{1}{2} {x}^{2} {y}^{2} = C$

Dec 11, 2017

Multiplying by $x$ we get an exact equation:

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

Whose solution is:

$y = \pm \frac{C}{x}$

ie a family of hyperbolas.

#### Explanation:

${y}^{2} \setminus \mathrm{dx} + x y \setminus \mathrm{dy} = 0 \ldots . . \left[A\right]$

Suppose we have:

$M \left(x , y\right) \setminus \mathrm{dx} + N \left(x , y\right) \setminus \mathrm{dy} = 0$

Then the DE is exact if ${M}_{y} - {N}_{x} = 0$

$M = {y}^{2} \implies {M}_{y} = 2 y$
$N = x y \implies {N}_{x} = y$

${M}_{y} - {N}_{x} \ne 0 \implies$ Not an exact DE

So, we seek an Integrating Factor $\mu \left(u\right)$ such that

${\left(\mu M\right)}_{y} = {\left(\mu N\right)}_{x}$

So, we compute::

$I = \frac{{M}_{y} - {N}_{x}}{N} = \frac{2 y - y}{x y} = \frac{1}{x}$

So the Integrating Factor is given by:

$\mu \left(x\right) = {e}^{\int \setminus I \setminus \mathrm{dx}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{\int \setminus \frac{1}{x} \setminus \mathrm{dx}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{\ln x}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = x$

So when we multiply the DE [A] by the IF we now get an exact equation:

$\left(x\right) {y}^{2} \setminus \mathrm{dx} + \left(x\right) x y \setminus \mathrm{dy} = 0 .$

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

And so if we redefine the function $M$ and $N$:

$M = x {y}^{2} \implies {M}_{y} = 2 x y$
$N = {x}^{2} y \implies {N}_{x} = 2 x y$

Making the equation exact

Then, our solution is given by:

${f}_{x} = M$ and ${f}_{y} = N$ and

If we consider ${f}_{y} = N$, then:

$f = \int \setminus {x}^{2} y \setminus \partial y + g \left(x\right)$, where we treat $x$ as constant
$\setminus \setminus = \frac{{x}^{2} {y}^{2}}{2} + g \left(x\right)$

And now we consider ${f}_{x} = M$ and we differentiate the last result to get:

${f}_{x} = x {y}^{2} + g ' \left(x\right)$

As ${f}_{x} = M$ then we have:

$x {y}^{2} + g ' \left(x\right) = x {y}^{2}$
$\therefore g ' \left(x\right) = 0 \implies g \left(x\right) = K$

$\frac{{x}^{2} {y}^{2}}{2} = K \implies {\left(x y\right)}^{2} = c$
$\therefore x y = \pm C \implies y = \pm \frac{C}{x}$