integrate y^′ −2ty +ty^2 =0


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Question Number 30421 by abdo imad last updated on 22/Feb/18

$i n t e g r a t e y^{^{'}} - 2 t y + {t y}^{2} = 0$
	Answered by sma3l2996 last updated on 24/Feb/18

	$y^{'} = t (2 y - y^{2})$ $\frac{d y}{y (2 - y)} = t d t$ $\frac{1}{y (2 - y)} = \frac{1}{2} (\frac{1}{y} + \frac{1}{2 - y})$ $\int \frac{d y}{y (2 - y)} = \int (\frac{1}{y} + \frac{1}{2 - y}) d y = l n ∣ y ∣ - l n ∣ 2 - y ∣= \frac{1}{2} t^{2} + λ$ $l n ∣ \frac{y}{2 - y} ∣= \frac{1}{2} t^{2} + λ$ $\frac{y}{2 - y} = e^{λ} \times e^{t^{2} / 2}$ $y = \times {K e}^{t^{2} / 2} (2 - y)$ $y (1 + {K e}^{t^{2} / 2}) = {K e}^{t^{2} / 2}$ $y (x) = \frac{{K e}^{t^{2} / 2}}{1 + {K e}^{t^{2} / 2}}$
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