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Integrating Factor

If for x ≥ 0,...

Question

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0,$ then y(3) is equal to

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Solution

$(1 + x) \frac{d y}{d x} = [(1 + x)^{2} + (y - 3)]$
$\Rightarrow (1 + x) \frac{d y}{d x} - y = (1 + x)^{2} - 3$
$\Rightarrow \frac{d y}{d x} - \frac{1}{(1 + x)} y = 1 + x - \frac{3}{1 + x}$

$I . F = e^{- \int \frac{1}{1 + x} d x} = \frac{1}{1 + x}$

$\Rightarrow y \times \frac{1}{1 + x} = \int 1 - \frac{3}{(1 + x)^{2}} d x$
$\Rightarrow \frac{y}{1 + x} = x + \frac{3}{1 + x} + c$
$\Rightarrow y = x (1 + x) + 3 + c (1 + x)$

At $x = 2, y = 0,$ we get
$0 = 6 + 3 + 3 c$
$\Rightarrow c = - 3$
At $x = 3,$
$y = x^{2} - 2 x = 9 - 6 = 3 \Rightarrow y (3) = 3$

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