QJEEM20Maaef4: JEE Main 2020 Mathematics - Class 12 JEEM20Maaef4 Solution: If for x 0, y = y(x) is the solution of the differential…

(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