# What is the function y if \frac { d y } { d x } = \frac { 3x ^ { 2} + 4x + 2} { 2( y - 1) } and y ( 0) = - 1?

Feb 2, 2017

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

#### Explanation:

$\setminus \frac{d y}{d x} = \setminus \frac{3 {x}^{2} + 4 x + 2}{2 \left(y - 1\right)}$

This is separable!

$\int 2 \left(y - 1\right) d y = \int 3 {x}^{2} + 4 x + 2 d x$

${y}^{2} - 2 y = {x}^{3} + 2 {x}^{2} + 2 x + C$

using the IV:

${\left(- 1\right)}^{2} - 2 \left(- 1\right) = C \implies C = 3$

$\implies {y}^{2} - 2 y - \left({x}^{3} + 2 {x}^{2} + 2 x + 3\right) = 0$

$y = \frac{2 \pm \sqrt{4 + 4 \left({x}^{3} + 2 {x}^{2} + 2 x + 3\right)}}{2}$

$y = 1 \pm \sqrt{{x}^{3} + 2 {x}^{2} + 2 x + 4}$

Checking the IV again:

$- 1 = 1 \pm \sqrt{0 + 4}$

And so the solution is the negative radical ie

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