# How do you solve y'=(2x-3)/(y^2+y+4) given y(1)=2?

Dec 13, 2016

$2 {y}^{3} + 3 {y}^{2} + 24 y = 6 {x}^{2} - 18 x + 44$

#### Explanation:

We have $y ' = \frac{2 x - 3}{{y}^{2} + y + 4}$, or

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

Which is a simple First Order separable DE, we can therefore separate the variables to get:
$\left({y}^{2} + y + 4\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x - 3\right)$
$\therefore \int \left({y}^{2} + y + 4\right) \mathrm{dy} = \int \left(2 x - 3\right) \mathrm{dx}$

We can integrate this to get:

$\frac{1}{3} {y}^{3} + \frac{1}{2} {y}^{2} + 4 y = {x}^{2} - 3 x + C$

Given $y \left(1\right) = 2$ we get:

$\frac{1}{3} {2}^{3} + \frac{1}{2} {2}^{2} + 4 \cdot 2 = {1}^{2} - 3 + C$
$\therefore \frac{8}{3} + 2 + 8 = 1 - 3 + C$
$\therefore C = \frac{44}{3}$

So the solution is:

$\frac{1}{3} {y}^{3} + \frac{1}{2} {y}^{2} + 4 y = {x}^{2} - 3 x + \frac{44}{3}$
$\therefore 2 {y}^{3} + 3 {y}^{2} + 24 y = 6 {x}^{2} - 18 x + 44$