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Question-101362




Question Number 101362 by bobhans last updated on 02/Jul/20
Commented by bemath last updated on 02/Jul/20
(x^2 +3y^2 )dx+(2xy−x)dy =0  M= x^2 +3y^2 → (∂M/∂y) = 6y  N = 2xy−x→(∂N/∂x) = 2y−1  (∂M/∂y) ≠ (∂N/∂x) ⇐ non exact
$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{2xy}−\mathrm{x}\right)\mathrm{dy}\:=\mathrm{0} \\ $$$$\mathrm{M}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} \rightarrow\:\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:=\:\mathrm{6y} \\ $$$$\mathrm{N}\:=\:\mathrm{2xy}−\mathrm{x}\rightarrow\frac{\partial\mathrm{N}}{\partial\mathrm{x}}\:=\:\mathrm{2y}−\mathrm{1} \\ $$$$\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:\neq\:\frac{\partial\mathrm{N}}{\partial\mathrm{x}}\:\Leftarrow\:\mathrm{non}\:\mathrm{exact}\: \\ $$

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