(x + 1)dy = ((x + 1)² + y – 3)dx

$$ \Rightarrow $$ (1 + x)$${{dy} \over {dx}}$$ - y = (1 + x)² - 3

$$ \Rightarrow $$ $${{dy} \over {dx}} - {y \over {1 + x}} = \left( {1 + x} \right) - {3 \over {1 + x}}$$

I.F = $${e^{ - \int {{{dx} \over {1 + x}}} }}$$ = $${1 \over {1 + x}}$$

Solution of the differential equation,

$$y\left( {{1 \over {1 + x}}} \right)$$ = $$\int {\left( {\left( {1 + x} \right) - {3 \over {1 + x}}} \right)\left( {{1 \over {1 + x}}} \right)dx} $$

$$ \Rightarrow $$ $${y \over {1 + x}}$$ = $$\int {{{{x^2} + 2x + 1 - 3} \over {{{\left( {x + 1} \right)}^2}}}dx} $$

$$ \Rightarrow $$ $${y \over {1 + x}}$$ = x + $${3 \over {1 + x}}$$ + C

As y(2) = 0 $$ \Rightarrow $$ x = 2, y = 0

$$ \therefore $$ 0 = 2 + $${3 \over {1 + 2}}$$ + C

$$ \Rightarrow $$ C = -3

So solution is $${y \over {1 + x}}$$ = x + $${3 \over {1 + x}}$$ - 3

y(3) means x = 3 and find value of y.

$${y \over {1 + 3}} = 3 + {3 \over {1 + 3}} - 3$$

$$ \Rightarrow $$ y = 3