# What is the general solution of the differential equation 2dy = 3xy \ dx?

Mar 25, 2017

$\text{ The G.S. is } 4 y = 3 {x}^{2} + c .$

#### Explanation:

$2 \mathrm{dy} = 3 x y \mathrm{dx} \Rightarrow 2 \frac{\mathrm{dy}}{y} = 3 x \mathrm{dx} ,$ which is a Separable Variable

Type Diff. Eqn.

To find its General Soln. (G.S.), we integrate it term-wise,

$\int 2 \mathrm{dy} = \int 3 x \mathrm{dx} + C .$

$\therefore 2 y = 3 \left({x}^{2} / 2\right) + C , \mathmr{and} , 4 y = 3 {x}^{2} + c , c = 2 C .$

Mar 25, 2017

$y = A {e}^{\frac{3}{4} {x}^{2}}$

#### Explanation:

We have:

$2 \mathrm{dy} = 3 x y \setminus \mathrm{dx}$

We should really write this as (because $\frac{d}{\mathrm{dx}}$ is an operator not a fraction):

$2 \frac{\mathrm{dy}}{\mathrm{dx}} = 3 x y$

Method 1 - Separating the Variables

If collect terms in $y$ and $x$ respectively:

$\frac{2}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 3 x$

And then we "separate the variables" to get:

$\int \setminus \frac{2}{y} \setminus \mathrm{dy} = \int \setminus 3 x \setminus \mathrm{dx}$

Integrating we get:

$\setminus \setminus \setminus 2 \ln y = \frac{3 {x}^{2}}{2} + K$
$\therefore \ln y = 3 {x}^{4} + \frac{1}{2} K$
$\therefore \setminus \setminus \setminus \setminus y = {e}^{\frac{3}{4} {x}^{2} + \frac{1}{2} K}$
$\therefore \setminus \setminus \setminus \setminus y = {e}^{\frac{3}{4} {x}^{2}} {e}^{\frac{1}{2} K}$
$\therefore \setminus \setminus \setminus \setminus y = A {e}^{\frac{3}{4} {x}^{2}}$

Method 2 - Integrating Factor

W can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

So:

$2 \frac{\mathrm{dy}}{\mathrm{dx}} = 3 x y$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{3}{2} x y = 0$ ..... [1]

Then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus - \frac{3}{2} x \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(- \frac{3}{4} {x}^{2}\right)$
$\setminus \setminus = {e}^{- \frac{3}{4} {x}^{2}}$

And if we multiply the DE [1] by this Integrating Factor, $I$, we will have a perfect product differential;

$\frac{\mathrm{dy}}{\mathrm{dx}} - \frac{3}{2} x y = 0$

$\therefore {e}^{- \frac{3}{4} {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{3}{2} x {e}^{- \frac{3}{4} {x}^{2}} y = 0$

$\therefore \frac{d}{\mathrm{dx}} \left\{{e}^{- \frac{3}{4} {x}^{2}} y\right\} = 0$

Which we can directly integrate to get:

${e}^{- \frac{3}{4} {x}^{2}} y = A$

And multiplying by ${e}^{\frac{3}{4} {x}^{2}}$ gives:

$y = {A}^{\frac{3}{4} {x}^{2}}$, as above