Show that the equation of the curve whose slope at any point is equal to $y + 2 x$ and which passes through the origin is $y = 2 (e^{x} - x - 1)$ .

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Solution

$\frac{d y}{d x} = y + 2 x \Rightarrow \frac{d y}{d x} - y = 2 x$
This is of the form $\frac{d y}{d x} + p y = Q$
Where $p = - 1, Q = 2 x$
$\int p d x = \int - 1 d x = - x$
$I F = e^{\int p d x} = e^{- x}$
Solution is $y (I F) = \int Q (I F) d x + c$
(i.e.,) $y e^{- x} = \int 2 x e^{- x} d x + c$
i.e.,) $y e^{- x} = 2 \int x d (e^{- x}) + c$
$y e^{- x} = - 2 x e^{- x} + 2 \frac{e^{- x}}{- 1} + c$
(i.e.,) $y e^{- x} = - 2 x e^{- x} - 2 e^{- x} + c$
Given, $y = 0$ when $x = 0$
$\Rightarrow$ $0 = 0 - 2 + c \Rightarrow c = 2$
$∴$ equation of the curve is
$y e^{- x} = - 2 x e^{- x} - 2 e^{- x} + 2$
(i.e.,) $y e^{- x} = 2 (1 - e^{- x} - x e^{- x})$
$\Rightarrow y = 2 e^{x} (1 - e^{- x} - x e^{- x})$
$\Rightarrow y = 2 (e^{x} - 1 - x)$
$\Rightarrow y = 2 (e^{x} - x - 1)$

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