Let y = y ( x ) be the solution of the differential equation x ^ 3...

Given differential equation is

${x}^{3}dy + \left(xy - 1\right)dx = 0$

$⇒ \dfrac{dy}{dx} = \dfrac{1 - xy}{{x}^{3}}$

$⇒ \dfrac{dy}{dx} = \dfrac{1}{{x}^{3}} - \dfrac{y}{{x}^{2}}$

$⇒ \dfrac{dy}{dx} + \dfrac{y}{{x}^{2}} = \dfrac{1}{{x}^{3}}$

This is a linear differential equation.

If $IF = {e}^{∫ \dfrac{1}{{x}^{2}}dx} = {e}^{- \dfrac{1}{x}}$

Solution of a given linear differential equation is

$y{e}^{- \dfrac{1}{x}} = ∫ {e}^{- \dfrac{1}{x}}\left(\dfrac{1}{{x}^{3}}\right)dx$

Put $- \dfrac{1}{x} = t ⇒ \dfrac{dx}{{x}^{2}} = dt$

$⇒ y{e}^{- \dfrac{1}{x}} = - ∫ {e}^{- t}tdt$

$⇒ y{e}^{- \dfrac{1}{x}} = t{e}^{- t} + {e}^{- t} + C$

$⇒ y = t + 1 + Ce\dfrac{1}{x}$

$⇒ y = \dfrac{1}{x} + 1 + Ce\dfrac{1}{x}$

Put $x = \dfrac{1}{2}$, then

$3 - e = 2 + 1 + C{e}^{2}$

$C = - \dfrac{1}{e}$

So,

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

So,

$y\left(1\right) = 1 + 1 - 1 = 1$