find critical point from Differential Equation System (dy/dx) = x − y (dy/dx) = x^2 +y^2 − 2


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Question Number 122732 by fajri last updated on 19/Nov/20

$f i n d c r i t i c a l p o i n t f r o m$ $D i f f e r e n t i a l E q u a t i o n S y s t e m$ $\frac{d y}{d x} = x - y$ $\frac{d y}{d x} = x^{2} + y^{2} - 2$
	Answered by bemath last updated on 19/Nov/20

	$(1) l e t y = x - u \Rightarrow \frac{d y}{d x} = 1 - \frac{d u}{d x}$ $\Rightarrow 1 - \frac{d u}{d x} = x - (x - u)$ $\Rightarrow 1 - u = \frac{d u}{d x}; \frac{d u}{1 - u} = d x$ $\Rightarrow \ln ∣ 1 - u ∣ = x + c$ $\Rightarrow 1 - u = \pm {K e}^{x}; 1 - (x - y) = \pm {K e}^{x}$ $\Rightarrow y = \pm {K e}^{x} + x - 1$ $\Rightarrow y^{'} = \pm {K e}^{x} + 1 = 0;$ $\Rightarrow e^{x} = \frac{1}{K}, w h e r e \frac{1}{K} > 0$ $\Rightarrow x = - \ln K a n d y = - {K e}^{- \ln K} - \ln K + 1$ $y = - K (\frac{1}{K}) - \ln K + 1 = - \ln K$ $H e n c e c r i t i c a l p o i n t o f y = f (x) = - {K e}^{x} + x - 1$ $i s (- \ln K, - \ln K) .$
	Commented by fajri last updated on 19/Nov/20

	$t h a n k S i r, I L i k e i t :)$
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