If x = x (t) is the solution of the differential equation (t + 1)dx...

Given: $\left(t + 1\right)dx = \left(2x + {\left(t + 1\right)}^{4}\right)dt$

$⇒ \dfrac{dx}{dt} = \dfrac{2x + {\left(t + 1\right)}^{4}}{\left(t + 1\right)}$

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

$⇒ IF = {e}^{∫ \dfrac{- 2}{t + 1}dt}$

$⇒ IF = {e}^{- 2\log \left|t + 1\right|}$

$⇒ IF = \dfrac{1}{{\left(t + 1\right)}^{2}}$

Solution of the differential equations is given by,

$⇒ x \times \dfrac{1}{{\left(t + 1\right)}^{2}} = ∫ \left(t + 1\right)dt$

$⇒ \dfrac{x}{{\left(t + 1\right)}^{2}} = \dfrac{{t}^{2}}{2} + t + c$

It is given that $x\left(0\right) = 2$

$⇒ \dfrac{2}{{\left(0 + 1\right)}^{2}} = \dfrac{0}{2} + 0 + c$

$⇒ c = 2$

$⇒ \dfrac{x}{{\left(t + 1\right)}^{2}} = \dfrac{{t}^{2}}{2} + t + 2$

Putting $t = 1$,

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

$⇒ \dfrac{x}{4} = \dfrac{1}{2} + 3$

$⇒ x = 2 + 12$

$⇒ x = 14$