# What is the solution of the Differential Equation  x(dy/dx)+3y+2x^2=x^3+4x ?

Jun 8, 2017

$y = \frac{5 {x}^{6} - 12 {x}^{5} + 30 {x}^{4} + C}{30 {x}^{3}}$

#### Explanation:

We have:

$x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 3 y + 2 {x}^{2} = {x}^{3} + 4 x$

We 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 rewrite the equations in standard form as:

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{3}{x} y = {x}^{2} - 2 x + 4 \ldots . . \left[1\right]$

Then the integrating factor is given by;

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

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

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) {x}^{3} + 3 {x}^{2} y = {x}^{5} - 2 {x}^{4} + 4 {x}^{3}$

$\therefore \frac{d}{\mathrm{dx}} \left({x}^{3} y\right) = {x}^{5} - 2 {x}^{4} + 4 {x}^{3}$

Which we can directly integrate to get:

${x}^{3} y = \int \setminus {x}^{5} - 2 {x}^{4} + 4 {x}^{3} \setminus \mathrm{dx}$

$\therefore {x}^{3} y = {x}^{6} / 6 - \frac{2 {x}^{5}}{5} + {x}^{4} + c$

$\therefore {x}^{3} y = \frac{5 {x}^{6} - 12 {x}^{5} + 30 {x}^{4} + 30 c}{30}$

$\therefore y = \frac{5 {x}^{6} - 12 {x}^{5} + 30 {x}^{4} + C}{30 {x}^{3}}$