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Question Number 168518 by LEKOUMA last updated on 12/Apr/22

Re^� soudre l′e^� quation aux differentielles  totales  2xydx+(x^2 −y^2 )dy=0

$${R}\acute {{e}soudre}\:{l}'\acute {{e}quation}\:{au}\mathrm{x}\:\mathrm{differentiel}{les} \\ $$$$\mathrm{totales} \\ $$$$\mathrm{2}{xydx}+\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$

Commented by mokys last updated on 12/Apr/22

M(x,y)=2xy → M_y  = 2x  N(x,y)=x^2 −y^2 →N_x = 2x    M_y  = N_x →the equation is exact    f = ∫ M(x,y)dx = ∫ (2xy)dx = x^2 y+A(y)    ∵ f_y  = N    ∴ x^2  + A^′ (y) = x^2 −y^2 ⇒A^′ (y)=−y^2 ⇒A(y)=−(y^3 /3) + C    ∴ x^2 y − (y^3 /3) = C     (aldolaimy)

$${M}\left({x},{y}\right)=\mathrm{2}{xy}\:\rightarrow\:{M}_{{y}} \:=\:\mathrm{2}{x} \\ $$$${N}\left({x},{y}\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} \rightarrow{N}_{{x}} =\:\mathrm{2}{x} \\ $$$$ \\ $$$${M}_{{y}} \:=\:{N}_{{x}} \rightarrow{the}\:{equation}\:{is}\:{exact} \\ $$$$ \\ $$$${f}\:=\:\int\:{M}\left({x},{y}\right){dx}\:=\:\int\:\left(\mathrm{2}{xy}\right){dx}\:=\:{x}^{\mathrm{2}} {y}+{A}\left({y}\right) \\ $$$$ \\ $$$$\because\:{f}_{{y}} \:=\:{N} \\ $$$$ \\ $$$$\therefore\:{x}^{\mathrm{2}} \:+\:{A}\:^{'} \left({y}\right)\:=\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \Rightarrow{A}\:^{'} \left({y}\right)=−{y}^{\mathrm{2}} \Rightarrow{A}\left({y}\right)=−\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\:+\:{C} \\ $$$$ \\ $$$$\therefore\:{x}^{\mathrm{2}} {y}\:−\:\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\:=\:{C}\: \\ $$$$ \\ $$$$\left({aldolaimy}\right) \\ $$

Answered by alephzero last updated on 12/Apr/22

please, translate to english

$${please},\:{translate}\:{to}\:{english} \\ $$

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