If (dx)/(dy) = 1 + x - y ^ 2 y , x(1) = 1, then 5x(2) is equal to:

Given: $\dfrac{dx}{dy} = \dfrac{1 + x - {y}^{2}}{y}$

$⇒ \dfrac{dx}{dy} = \dfrac{1 + x}{y} - y$

$⇒ \dfrac{dx}{dy} - \dfrac{1 + x}{y} = - y$

Now let $1 + x = t ⇒ \dfrac{dx}{dy} = \dfrac{dt}{dy}$

$⇒ \dfrac{dt}{dy} - \dfrac{t}{y} = - y$

$⇒ IF = {e}^{∫ - \dfrac{1}{y}dy}$

$⇒ IF = {e}^{- \log y}$

$⇒ IF = {e}^{log\dfrac{1}{y}}$

$⇒ IF = \dfrac{1}{y}$

So, the solution of the given differential equation is

$\dfrac{t}{y} = ∫ \dfrac{1}{y} \times \left(- y\right)dy$

$⇒ \dfrac{\left(1 + x\right)}{y} = ∫ \dfrac{1}{y} \times \left(- y\right)dy$

$⇒ \dfrac{\left(1 + x\right)}{y} = - y + c$

Also, $x\left(1\right) = 1$

$⇒ \dfrac{2}{1} = - 1 + c$

$⇒ c = 3$

$⇒ \dfrac{\left(1 + x\right)}{y} = - y + 3$

Putting $y = 2$

$⇒ \dfrac{\left(1 + x\right)}{2} = - 2 + 3$

$⇒ \left(1 + x\right) = 2$

$⇒ x = 1$

$⇒ 5x = 5$