# Question #3e972

Feb 8, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y + 1 + {x}^{2}}{x - {x}^{2} \left(\sin y\right)}$

#### Explanation:

$y \mathrm{dx} - x \mathrm{dy} + \left(1 + {x}^{2}\right) \mathrm{dx} + {x}^{2} \left(\sin y\right) \mathrm{dy} = 0$

$y \mathrm{dx} + \left(1 + {x}^{2}\right) \mathrm{dx} = x \mathrm{dy} - {x}^{2} \left(\sin y\right) \mathrm{dy}$

$\mathrm{dx} \left[y + 1 + {x}^{2}\right] = \mathrm{dy} \left[x - {x}^{2} \left(\sin y\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y + 1 + {x}^{2}}{x - {x}^{2} \left(\sin y\right)}$

Feb 8, 2018

$y = x \left(x - \cos y + C\right) - 1$.

#### Explanation:

Let us rewrite the given Diff. Eqn. as,

$\left(1 + {x}^{2}\right) \mathrm{dx} + {x}^{2} \sin y \mathrm{dy} = x \mathrm{dy} - y \mathrm{dx}$.

Dividing by ${x}^{2}$, we have,

$\left(\frac{1 + {x}^{2}}{x} ^ 2\right) \mathrm{dx} + \sin y \mathrm{dy} = \frac{x \mathrm{dy} - y \mathrm{dx}}{x} ^ 2$.

The Right Member of the eqn. is note-worthy :

Observe that, $d \left(\frac{y}{x}\right) = \frac{x \mathrm{dy} - y \mathrm{dx}}{x} ^ 2$.

Utilising this, we find that the given eqn. is,

$\left(\frac{1 + {x}^{2}}{x} ^ 2\right) \mathrm{dx} + \sin y \mathrm{dy} = d \left(\frac{y}{x}\right)$, which looks like,

separable variable type. So, Integrating term-wise,

$\int \left(\frac{1 + {x}^{2}}{x} ^ 2\right) \mathrm{dx} + \int \sin y \mathrm{dy} + C = \int d \left(\frac{y}{x}\right)$.

$\therefore \int \left(\frac{1}{x} ^ 2 + 1\right) \mathrm{dx} - \cos y + C = \frac{y}{x} , i . e . ,$

$- \frac{1}{x} + x - \cos y + C = \frac{y}{x} , \mathmr{and} ,$

$y = x \left(x - \cos y + C\right) - 1 ,$ is the desired General Solution!